Edward  Bright 


61 R 


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XK 


THE  : h- 


*         *  «  *  9    * 


ESSENTIALS  OF  GEOMETRY 


BY 


WEBSTER   WELLS,   S.B. 

//     .  . 

PROFESSOR  OF  MATHEMATICS  IN    THE   MASSACHUSETTS 
INSTITUTE  OF  TECHNOLOGY 


3>*:c 


BOSTON,  U.S.A. 
D.  C.   HEATH  &  CO.,  PUBLISHERS 

1899 


Copyright,  1898  and  1899, 
By  WEBSTER  WELLS. 


Norfajootr  i^ress 

J.  S.  Gushing  &  Co.  -  Berwick  &  Smith 
Norwood  Mass.  U.S.A. 


.3<^ 


PREFACE. 


In  the  Essentials  of  Geometry,  the  author  has  endeavored 
to  prepare  a  work  suited  to  the  needs  of  high  schools  and 
academies.  It  will  also  be  found  to  answer  as  well  the 
requirements  of  colleges  and  scientific  schools. 

In  some  of  its  features,  the  work  is  similar  to  the  author's 
Revised  Plane  and  Solid  Geometry;  but  important  improve- 
ments have  been  introduced,  which  are  in  line  with  the 
present  requirements  of  many  progressive  teachers. 

In  a  number  of  propositions,  the  figure  is  given,  and  a 
statement  of  what  is  to  be  proved ;  the  details  of  the  proof 
being  left  to  the  pupil,  usually  with  a  hint  as  to  the  method 
of  demonstration  to  be  employed. 

The  propositions  and  corollaries  left  in  this  way  for  the 
pupil  to  demonstrate,  in  the  Plane  Geometry,  will  be  found 
in  the  following  sections  :  — 

Book  I.,  §§  51,  75,  76,  78,  79,  96,  102,  110,  111,  112,  115, 
117, 136. 

Book  II.,  §§  158,  160,  165, 170, 172  (Case  III.),  174, 178, 
179, 193  (Case  III.),  194,  and  201. 

Book  III.,  §§  251,  257,  261,  264,  268,  278,  282,  284,  and 
286. 

Book  IV.,  §§  312  and  316. 

Book  v.,  §§  346,  347,  and  350. 

iii 

797944 


iy  PREFACE. 

Book  VI.,  §§  405,  407,  412,  414,  415,  416,  417,  420,  421, 
434,  437,  440,  442,  and  444. 

Book  VII.,  §§  491,  495,  507,  512,  513,  521,  528,  529,  and 
530. 

Book  VIII.,  §§  554,  559,  578,  580,  581,  594,  595,  601,  603, 
608,  613,  614,  625  (Case  II.),  630,  631,  635,  and  637. 

Book  IX.,  §§  654,  656,  660,  673,  and  679. 

There  are  also  Problems  in  Construction  in  which  the 
construction  or  proof  is  left  to  the  pupil. 

Another  important  improvement  consists  in  giving  figures 
and  suggestions  for  the  exercises.  In  Book  I.,  the  pupil 
has  a  ligure  for  every  non-numerical  exercise;  after  that, 
they  are  only  given  with  the  more  difficult  ones. 

In  many  of  the  exercises  in  construction,  the  pupil  is 
expected  to  discuss  the  problem,  or  point  out  its  limitations. 

In  Book  I.,  and  also  in  the  first  eighteen  propositions  of 
Book  VI.,  the  authority  for  each  statement  of  a  proof  is 
given  directly  after  the  statement,  in  smaller  type,  enclosed 
in  brackets.  In  the  remaining  portions  of  the  work,  the 
formal  statement  of  the  authority  is  omitted ;  but  the  num- 
ber of  the  section  where  it  is  to  be  found  is  usually  given. 

In  a  number  of  cases,  however,  where  the  pupil  is  pre- 
sumed, from  practice,  to  be  so  familiar  with  the  authority 
as  not  to  require  reference  to  the  section  where  it  is  to  be 
found,  there  is  given  merely  an  interrogation-point. 

In  all  these  cases  the  pupil  should  be  required  to  give 
the  authority  as  carefully  and  accurately  as  if  it  were  actu- 
ally printed  on  the  page. 

Another  improvement  consists  in  marking  the  parts  of 
a  demonstration  by  the  words  Given,  To  Pi'ove,  and  Proof, 
printed  in  heavy -faced  type. 

A  similar  system  is  followed  in  the  Constructions,  by  the 
use  of  the  words  Given,  Required,  Construction,  and  Proof. 

A  minor  improvement  is  the  omission  of  the  definite 
article  in  speaking  of  geometrical  magnitudes;  thus  we 
speak  of  "angle  ^,"  "triangle  ABC,^^  etc.,  and  not  "the 
angle  A,^^  "the  triangle  ABC,''  etc. 


PREFACE. 

Symbols  and  abbreviations  have  been  freely  used ;  a  list 
of  these  will  be  found  on  page  4. 

Particular  attention  has  been  given  to  putting  the  propo- 
sitions in  the  first  part  of  Book  I.  in  a  form  adapted  to  the 
needs  of  a  beginner. 

The  pages  have  been  arranged  in  such  a  way  as  to  avoid 
the  necessity,  while  reading  a  proof,  of  turning  the  page  for 
reference  to  the  figure. 

The  Appendix  to  the  Plane  Geometry  contains  proposi- 
tions on  Maxima  and  Minima  of  Plane  Figures,  and  Sym- 
metrical Figures;  also,  additional  exercises  of  somewhat 
greater  difficulty  than  those  previously  given. 

The  Appendix  to  the  Solid  Geometry  contains  rigorous 
proofs  of  the  limit  statements  made  in  §§  639,  650,  667, 
and  674. 

The  author  wishes  to  acknowledge,  with  thanks,  the 
many  suggestions  which  he  has  received  from  teachers  in 
all  parts  of  the  country,  which  have  added  materially  to 
the  value  of  the  work. 

WEBSTER  WELLS. 

Massachusetts  Institute  of  Technology, 


Stereoscopic  views  of  many  of  the  figures  in  the  Solid  Geometry 
have  been  prepared.  Full  particulars  may  be  obtained  from  the 
publishers. 


Digitized  by  tine  Internet  Arciiive 

in  2007  witii  funding  from 

IVIicrosoft  Corporation 


littp://www.arcliive.org/details/essentialsofgeomOOwellricli 


CONTENTS. 


PAGE 

Preliminary  Definitions    .        .        .        .        .        .        .        i 

PLANE  GEOMETRY. 

Book  1.  Rectilinear  Figures 5 

Book  IL  The  Circle 72 

Book  III.  Theory  of  Proportion.  —  Similar  Polygons  122 

Book  IV.  Areas  of  Polygons 162 

Book       V.     Regular  Polygons.  —  Measurement  of  the 

Circle .        .188 

APPENDIX  TO  PLANE   GEOMETRY. 

Maxima  and  Minima  of  Plane  Figures  .        .        .    211 

Symmetrical  Figures 217 

Additional  Exercises 220 

SOLID   GEOMETRY. 

Book      VI.     Lines    and    Planes    in    Space.  —  Diedral 

Angles.  —  Polyedral  Angles  .        .'        .    233 

Book    VII.     Polyedrons 273 

Book  VIII.     The  Cylinder,  Cone,  and  Sphere       .        .    319 

Book      IX.    Measurement  of  the  Cylinder,  Cone,  and 

Sphere '  •        •        •    360 

Appendix  to  Solid  Geometry 386 


GEOMETRY. 


>>»i< 


PRELIMINARY  DEFINITIONS. 


(^ 


A  material  body. 


A 
/\ 

/ 

/ 

A  geometrical  solid. 


1.  A  material  body,  such  as  a  block  of  wood,  occupies  a 
limited  or  hounded  portion  of  space. 

The  boundary  which  separates  such  a  body  from  sur- 
rounding space  is  called  the  surface  of  the  body. 

2.  If  the  material  composing  such  a  body  could  be  con- 
ceived as  taken  away  from  it,  without  altering  the  form  or 
shape  of  the  hounding  surface,  there  would  remain  a  portion 
of  space  having  the  same  bounding  surface  as  the  former 
material  body ;  this  portion  of  space  is  called  a  geometrical 
solid,  or  simply  a  solid. 

The  surface  which  bounds  it  is  called  a  geometrical  sur- 
face, or  simply  a  surface;  it  is  also  called  the  surface  of  the 
solid. 

3.  If  two  geometrical  surfaces  intersect 
each  other,  that  which  is  common  to  both  is 
called  a  geometriccd  line,  or  simply  a  line. 

Thus,  if  surfaces  ^B  and  CD  cut  each 
other,  their  common  intersection,  EF,  is  a 
line. 

1 


GEOMETRY. 


4'.   If    two  '  geometrical    lines    intersect    ^\  ^D 

each  other,  that  which  is  common  to  both 
is  called  a  geometrical  point,  or  simply  a 
point.  ^^  ^^ 

Thus,  if  lines  AB  and  CD  cut  each  other,  their  common 
intersection,  0,  is  a  point. 

5.  A  solid  has  extension  in  every  direction;  but  this  is  not 
true  of  surfaces  and  lines. 

A  point  has  extension  in  no  direction,  but  simply  position 
in  space. 

6.  A  surface  may  be  conceived  as  existing  independently 
in  space,  without  reference  to  the  solid  whose  boundary  it 
forms. 

In  like  manner,  we  may  conceive  of  lines  and  points  as 
having  an  independent  existence  in  space. 

7.  A  straight  line,  or  right  line,  is  a  line  which  has  the 
same  direction  throughout  its  length ;  as  AB. 

F        G 

A- 


A  curved  line,  or  curve,  is  a  line  no  portion  of  which  is 
straight;  as  CD. 

A  broken  line  is  a  line  which  is  composed  of  different 
successive  straight  lines ;  as  EFGH. 

8.  The  word  "  line  "  will  be  used  hereafter  as  signifying 
a  straight  line. 

9.  A  plane  surface,  or  plane,  is  a  surface  such  that  the 
straight  line  joining  any  two  of  its  points        jf 
lies  entirely  in  the  surface. 

Thus,  if  P  and  Q  are  any  two  points  in    /^ 
surface  MN,  and  the  straight  line  joining 
P  and  Q  lies  entirely  in  the  surface,  then  MN  is  a  plane. 

10.  A  curved  surface  is  a  surface  no  portion  of  which  is 
plane. 


PRELIMINARY  DEFINITIONS.  .3 

11.  We  may  conceive  of  a  straight  line  as  being  of  un- 
limited extent  in  regard  to  length;  and  in  like  manner  we 
may  conceive  of  a  plane  as  being  of  unlimited  extent  in 
regard  to  length  and  breadth. 

12.  A  geometrical  figure  is  any  combination  of  points, 
lines,  surfaces,  or  solids. 

A  plane  figure  is  a  figure  formed  by  points  and  lines  all 
lying  in  the  same  plane. 

A  geometrical  figure  is  called  rectilinear,  or  nght-Uned, 
when  it  is  composed  of  straight  lines  only. 

13.  Geometry  treats  of  the  properties,  construction,  and 
measurement  of  geometrical  figures. 

14.  Plane  Geometry  treats  of  plane  figures  only. 

Solid  Geometry,  also  called  Geometry  of  Space,  or  Geometry 
of  Three  Dimensions,  treats  of  fi.gures  which  are  not  plane. 

15.  An  Axiom  is  a  truth  which  is  assumed  without  proof 
as  being  self-evident. 

A  TJieorem  is  a  truth  which  requires  demonstration. 

A  Problem  is  a  question  proposed  for  solution. 

A  Proposition  is  a  general  term  for  a  theorem  or  problem. 

A  Postulate  assumes  the  possibility  of  solving  a  certain 
problem. 

A  Corollary  is  a  secondary  theorem,  which  is  an  imme- 
diate consequence  of  the  proposition  which  it  follows. 

A  Scholium  is  a  remark  or  note. 

An  Hypothesis  is  a  supposition  made  either  in  the  state- 
ment or  the  demonstration  of  a  proposition. 

16.  Postulates. 

1.  We  assume  that  a  straight  line  can  be  drawn  between 
any  two  points. 

2.  We  assume  that  a  straight  line  can  be  produced  (i.e., 
prolonged)  indefinitely  in  either  direction. 

17.  Axioms. 

We  assume  the  truth  of  the  following: 


4  GEOMETRY. 

1.  Things  which  are  equal  to  the  same  thing,  or  to  equals, 
are  equal  to  each  other. 

2.  If  the  same  operation  he  performed  upon  equals,  the 
results  will  be  equal. 

3.  Bid  one  straight  line  can  be  drawn  between  two  points. 

4.  A  straight  line  is  the  shortest  line  between  two  points. 

5.  The  whole  is  equal  to  the  sum  of  all  its  parts. 

6.  The  whole  is  greater  than  any  of  its  parts. 


18.   Since  but  one  straight 

line  can  be  drawn  between  two 

points,  a  straight  line  is  said  to  be  determined  by  any  two  of 

its  points. 

19.  Symbols  and  Abbreviations. 

The  following  symbols  will  be  used  in  the  work : 

+,  plus. 

A,  triangle. 

— ,  minus. 

A,  triangles. 

X ,  multiplied  by. 

J_,  perpendicular,   is    perpen- 

=, equals. 

dicular  to. 

=c=,  equivalent,  is  equivalent 

Js,  perpendiculars. 

to. 

II  ,  parallel,  is  parallel  to. 

>,  is  greater  than. 

lis,  parallels. 

<,  is  less  than. 

O,  parallelogram. 

.-.,  therefore. 

m,  parallelograms. 

Z,  angle. 

O,  circle. 

A,  angles. 

CD,  circles. 

The  following  abbreviations  will  also  be  used : 

Ax.,     Axiom. 

Sup.,         Supplementary. 

Def.,    Definition. 

Alt.,         Alternate. 

Hyp.,  Hypothesis. 

Int.,          Interior. 

Cons.,  Construction. 

Ext.,         Exterior. 

Rt.,      Right. 

Corresp.,  Corresponding. 

Str.,     Straight. 

Rect.,       Rectangle,  rec- 

Adj.,   Adjacent. 

tangular. 

PLAICE    GEOMETRY, 


Book  I. 

RECTILINEAR  FIGURES. 


DEFINITIONS  AND   GENERAL  PRINCIPLES. 

20.  An  angle  (Z)  is  the  amount  of  diverg- 
ence of  two  straight  lines  which  are  drawn 
from  the  same  point  in  different  directions. 

The  point  is  called  the  vertex  of  the  angle, 
and  the  straight  lines  are  called  its  sides. 

21.  If  there  is  but  one  angle  at  a  given  vertex,  it  may  be 
designated  by  the  letter  at  that  vertex ;  but  if  two  or  more 
angles  have  the  same  vertex,  we  avoid  ambiguity  by  naming 
also  a  letter  on  each  side,  placing  the  letter  at  the  vertex 
between  the  others. 

Thus,  we  should  call  the  angle  of  §  20  "angle  0";  but 
if  there  were  other  angles  having  the  same  vertex,  we 
should  read  it  either  AOB  or  BOA. 

Another  way  of  designating  an  angle  is  by  means  of  a 
letter  placed  between  its  sides;  examples  of  this  will  be 
found  in  §  71. 

22.  Two  geometrical  figures  are  said  to  be  equal  when 
one  can  be  applied  to  the  other  so  that  they  shall  coincide 
throughout. 

To  prove  two  angles  equal,  we  do  not  consider  the  lengths 
of  their  sides. 

5 


6 


PLANE   GEOMETRY.— BOOK  I. 


Thus,  if  angle  ABC  can  be  applied  to  angle  DEF  in  such 
a  manner  that  point  B  shall  fall 
on  point  E,  and  sides  AB  and 
BG  on  sides  DE  and  EF,  respec- 
tively, the  angles  are  equal,  even 
if  sides  AB  and  BC  are  not  equal 
in  length  to  sides  DE  and  EF,  respectively. 

23.  Two  angles  are  said  to  be  adjacent 
when  they  have  the  same  vertex,  and  a 
common  side  between  them;  as  AOB  and 

Boa  0^ 


PERPENDICULAR  LINES. 

24.  If  from  a  given  point  in  a  straight  line  a  line  be 
drawn  meeting  the  given  line  in  such  a  way  as  to  make  the 
adjacent  angles  equal,  each  of  the  equal  angles  is  called  a 
right  angle,  and  the  lines  are  said  to  be  perpendicular  (±) 
to  each  other.  ^ 

Thus,  if  from  point  A  in  straight  line  CD 
line  ABhe  drawn  in  such  a  way  as  to  make 
angles  BAC  and  BAD  equal,  each  of  these 
angles  is  a  right  angle,  and  AB  and  CD  are 
perpendicular  to  each  other.  A         ~ 

Pkop.  I.     Theorem. 

25.  At  a  given  point  in  a  straight  line,  a  perpendicular  to 
the  line  can  he  drawn,  and  hut  one. 


A  C  B 

Let  C  be  the  given  point  in  straight  line  AB. 


RECTILINEAR  FIGURES.  7 

To  prove  that  a  perpendicular  can  be  drawn  to  AB  at  C, 
and  but  one. 

Draw  a  straight  line  CD  in  such  a  position  that  angle 
BCD  shall  be  less  than  angle  ACD;  and  let  line  CD  be 
turned  about  point  C  as  a  pivot  towards  the  position  CA. 

Then,  angle  BCD  will  constantly  increase;  and  angle 
ACD  will  constantly  diminish,  until  it  becomes  less  than 
angle  BCD-,  and  it  is  evident  that  there  is  one  position 
of  CD,  and  only  one,  in  which  these  angles  are  equal. 

Let  CE  be  this  position ;  then  by  the  definition  of  §  24, 
CE  is  perpendicular  to  AB. 

Hence,  a  perpendicular  can  be  drawn  to  AB  at  C,  and 
but  one. 

26.  Cor.   All  right  angles  are  equal. 

Let  ABC  SindDEF  be  right  angles. 

To  prove  angles  ABC  and  DEF 
equal.  a 

Let  angle  ABC  be  superposed  (i.e., 
placed)  upon  angle  DEF  in  such  a  way  that  point  B  shall 
fall  upon  point  E,  and  line  AB  upon  line  DE. 

Then,  line  BC  will  fall  upon  line  EF;  for  otherwise  we 
should  have  two  lines  perpendicular  to  DE  at  E,  which  is 
impossible. 

[At  a  given  point  in  a  straight  line,  but  one  perpendicular  to  the 
line  can  be  drawn.  ]  (§25) 

Hence,  angles  ABC  and  DEF  are  equal  (§  22). 


B       D 


DEFINITIONS. 

27.  An  acute  angle  is  an  angle  which 
is  less  than  a  right  angle ;  as  ABC. 

An  obtuse  angle  is  an  angle  which  is 
greater  than  a  right  angle ;  as  DEF. 

Acute  and  obtuse  angles  are  called 
oblique  angles;  and  intersecting  lines 
which  are  not  perpendicular,  are  said  to 
be  oblique  to  each  other. 


8  PLANE   GEOMETRY. —BOOK  I. 

28.  Two  angles  are  said  to  be  veHical,    ^. 
or  opposite,  when  the  sides  of  one  are  the 
prolongations  of  the  sides  of  the  other ;  as 
AEC  and  BED. 

29.  An  angle  is  measured  by  finding  how  many  times  it 
contains  another  angle,  adopted  arbitrarily  as  the  unit  of 
measure. 

The  usual  unit  of  measure  is  the  degree,  which  is  the 
ninetieth  part  of  a  right  angle. 

To  express  fractional  parts  of  the  unit,  the  degree  is 
divided  into  sixty  equal  parts  called  minutes,  and  the  min- 
ute into  sixty  equal  parts,  called  seconds. 

Degrees,  minutes,  and  seconds  are  represented  by  the 
symbols,  °,  ',  ",  respectively. 

Thus,  43°  22'  37"  represents  an  angle  of  43  degrees,  22 
minutes,  and  37  seconds. 

30.  If  the  sum  of  two  angles  is  a  right  angle,  or  90°,  one 
is  called  the  compleynent  of  the  other;  and  if  their  sum  is 
two  right  angles,  or  180°,  one  is  called  the  supplement  of  the 
other. 

For  example,  the  complement  of  an  angle  of  34°  is 
90°  —  34°,  or  56° ;  and  the  supplement  of  an  angle  of  34°  is 
180°  -  34°,  or  146°. 

Two  angles  which  are  complements  of  each  other  are 
called  complementary;  and  two  angles  which  are  supple- 
ments of  each  other  are  called  supplementary. 

31.  It  is  evident  that 

1.  The  complements  of  equal  angles  are  equal. 

2.  The  supplements  of  equal  angles  are  equal. 

EXERCISES. 

1.  How  many  degrees  are  there  in  the  complement  of  47°  ?  of  83°  ? 
of  90°? 

2.  How  many  degrees  are  there  in  the  supplement  of  31°  ?  of  90°  ? 
of  178°  ? 


RECTILINEAR   FIGURES.  9 

3.  How  many  degrees  are  there  in  the  complement,  and  in  the 
supplement,  of  an  angle  equal  to  /j  of  a  right  angle  ? 

4.  How  many  degrees  are  there  in  an  angle  whose  supplement  is 
equal  to  f  f  of  its  complement  ? 

5.  Two  angles  are  complementary,  and  the  greater  exceeds  the 
less  by  37°.    How  many  degrees  are  there  in  each  angle  ? 

Prop.  II.     Theorem. 

32.   If  two  adjacent  angles  have  their  exterior  sides  in  the 
same  straight  line,  their  sum  is  equal  to  two  right  angles. 

E 


A  C  B 

Let  angles  ACD  and  BCD  have  their  sides  AC  and  BC 
in  the  same  straight  line. 

To  prove  the  sum  of  angles  ACD  and  BCD  equal  to  two 
right  angles. 

Draw  line  CE  perpendicular  to  AB  at  C. 

[At  a  given  point  in  a  straight  line,  a  perpendicular  to  the  line  can 
be  drawn.]  (§  25) 

Then,  it  is  evident  that  the  sum  of  angles  ACD  and  BCD 
is  equal  to  the  sum  of  angles  ACE  and  BCE. 

But  since  CE  is  perpendicular  to  AB,  angles  ACE  and 
BCE  are  right  angles. 

Hence,  the  sum  of  angles  ACD  and  BCD  is  equal  to  two 
right  angles. 

33.  Sch.  Since  angles  ACD  and  BCD  are  supplementary 
(§  30),  the  theorem  may  be  stated  as  follows : 

If  two  adjacent  angles  have  their  exterior  sides  in  the  same 
straight  line,  they  are  supplementary. 

Such  angles  are  called  supplementary-adjacent. 


10  PLANE   GEOMETRY.— BOOK  I. 

34.  Cor.  I.  TTie  sum  of  all  the  angles  on  the  same  side  of 
a  straight  line  at  a  given  point  is  equal  to  two  right  angles. 

This  is  evident  from  §  32. 

35.  Cor.  II.  Tfie  sum  of  all  the  angles  about  a  point  in 
a  plane  is  equal  to  four  right  angles. 

Let  AOB,  BOG,  COD,  and  DOA  be  angles  about  the 
point  0. 

To  prove  the  sum  of  angles  AOB,  \  _B 

BOG,  GOD,  and  DOA  equal  to  four  \      ./^ 

right  angles.  -E -\^- A 

Produce  ^0  to  ^.  / 

Then,  the  sura  of  angles  AOB,  BOG,  I 

and  GOE  is  equal  to  two  right  angles.  'D 

[The  sum  of  all  the  angles  on  the  same  side  of  a  straight  line  at 
a  given  point  is  equal  to  two  right  angles.]  (§  34) 

In  like  manner,  the  sum  of  angles  EOD  and  DOA  is 
equal  to  two  right  angles. 

Therefore,  the  sum  of  angles  AOB,  BOG,  GOD,  and  DOA 
is  equal  to  four  right  angles. 


Ex.  6.  If,  in  the  figure  of  §  35,  angles  AOB,  BOC,  and  COD  are 
respectively  49°,  88°,  and  |  of  a  right  angle,  how  many  degrees  are 
there  in  angle  AOD  ? 

36.  Sch.  The  pupil  will  now  observe  that  a  demonstrar 
tion,  in  Geometry,  consists  of  three  parts : 

1.  The  statement  of  what  is  given  in  the  figure. 

2.  The  statement  of  what  is  to  be  proved. 

3.  The  proof 

In  the  remaining  propositions  of  the  work,  we  shall  mark 
clearly  the  three  divisions  of  the  demonstration  by  heavy- 
faced  type,  and  employ  the  svmbols  and  abbreviations  of 
§  20. 


RECTILINEAR  FIGURES.  n 

Prop.  III.     Theorem. 

37.   If  the  sum  of  two  adjacent  angles  is  equal  to  two  right 
angles,  their  exterior  sides  lie  in  the  same  straight  line. 


.-"E 


A  C  B 

Given  the  sum  of  adj.  A  ACD  and  BCD  equal  to  two 

rt.  A. 

To  Prove  that  AC  and  BC  lie  in  the  same  str.  line. 

Proof.  If  AC  and  BC  do  not  lie  in  the  same  str.  line,  let 
CE  be  in  the  same  str.  line  with  AC. 

Then  since  ACE  is  a  str.  line,  Z  ECD  is  the  supplement 
of  A  ACD. 

[If  two  adj.  A  have  their  ext.  sides  in  the  same  str.  line,  they  are 
supplementary.]  (§33) 

But  by  hyp.,  A  ACD  +  A  BCD  =  two  rt.  A. 
Whence,  A  BCD  is  the  supplement  of  A  ACD.         (§  30) 
Then  since  both  A  ECD  and  A  BCD  are  supplements  of 
A  ACD,  A  ECD  =  A  BCD. 

[The  supplements  of  equal  A  are  equal.]  (§  31) 

Hence,  EC  coincides  with  BC,  and  AC  and  BC  lie  in  the 
same  str.  line. 

38.  Sch.  I.  It  will  be  observed  that  the  enunciation  of 
every  theorem  consists  essentially  of  two  parts ;  the  Hypoth- 
esis, and  the  Conclusion. 

Thus,  we  may  enunciate  Prop.  I  as  follows : 
Hypothesis.     If  a  point  be  taken  in  a  given  straight  line, 
Conclusion.     A  perpendicular  to  the  line  at  the  given  point 
can  be  drawn,  and  but  one. 


12  PLANE   GEOMETRY. —BOOK  I. 

39.  Sch.  II.     We  may  enunciate  Prop.  II  as  follows  : 
Hypothesis.     If  two  adjacent  angles  have  their  exterior 

sides  in  the  same  straight  line, 

Conclusion.     Their  sum  is  equal  to  two  right  angles. 

Again,  we  may  enunciate  Prop.  Ill : 

Hypothesis.  If  the  sum  of  two  adjacent  angles  is  equal 
to  two  right  angles, 

Conclusion.  Their  exterior  sides  lie  in  the  same  straight 
line. 

One  proposition  is  said  to  be  the  Converse  of  another  when 
the  hypothesis  and  conclusion  of  the  first  are,  respectively, 
the  conclusion  and  hypothesis  of  the  second. 

It  is  evident  from  the  above  considerations  that  Prop.  Ill 
is  the  converse  of  Prop.  II. 

Prop.  IV.     Theorem. 

40.  If  two  straight  lines  intersect,  the  vertical  angles  are 
equal. 


Given  str.  lines  AB  and  CD  intersecting  at  0. 
To  Prove  ZAOO  =  ZBOD. 

Proof.     Since  AAOC  and  AOD  have  their  ext.  sides  in 
str.  line  CD,  ZAOC  is  the  supplement  of  Z  AOD. 

[If  two  adj.  A  have  their  ext.  sides  in  the  same  str.  line,  they  are 
supplementary.]  (§  33) 

For  the  same  reason,  Z  BOD  is  the  supplement  of  Z  AOD, 

.-.  ZAOC=ZBOD. 

[The  supplements  of  equal  A  are  equal.]  (§  31) 

In  like  manner,  we  may  prove 

ZAOD  =  ZBOa 


RECTILINEAR  FIGURES. 


13 


EXERCISES. 

7.  If,  in  the  figure  of  Prop.  IV.,  ZAOD  =  137°,  how  many  degrees 
are  there  in  J50C?   in^OC?   in  BOD? 

8.  Two  angles  are  supplementary,  and  the  greater  is  seven  times 
the  less.    How  many  degrees  are  there  in  each  angle  ? 

Prop.  V.     Theorem. 

41.  If  a  perpendicular  be  erected  at  the  middle  point  of  a 
straight  line, 

I.   Any  point  in  the  perpendicular  is  equally  distant  from 
the  extremities  of  the  line. 

II.   Any  point  without  the  perpendicular  is  unequally  dis- 
tant from  the  extremities  of  the  line. 


I.  Given  line  CD  ±  to  line  AB  at  its  middle  point  D,  E 
any  point  in  CD,  and  lines  AE  and  BE. 

To  Prove  AE  =  BE. 

Proof.  Superpose  figure  BDE  upon  figure  ADE  by  fold- 
ing it  over  about  line  DE  as  an  axis. 

Now  Z  BDE  =  Z  ADE. 

[All  rt.  A  are  equal.]  •     (§  26) 

Then,  line  BD  will  fall  upon  line  AD. 

But  by  hyp.,  BD  =  AD. 

Whence,  point  B  will  fall  on  point  A. 

Then  line  BE  will  coincide  with  line  AE. 

[But  one  str.  line  can  be  drawn  between  two  points.]  (Ax.  3) 

.-.  AE  =  BE. 


14  PLANE   GEOMETRY.— BOOK  I. 


A  D  B 

II.  Given  line  CD  ±  to  line  AB  at  its  middle  point  D, 
F  any  point  without  CD,  and  lines  AF  and  BF. 

To  Prove  AF  >  BF, 

Proof.   Let  AF  intersect  CD  at  E,  and  draw  line  J5^. 

Now  BE  +  EF>BF. 

[A  str.  line  is  the  shortest  line  between  two  points.]  (Ax.  4) 

But,  BE  =  AE. 

[If  a  _L  be  erected  at  the  middle  point  of  a  str.  line,  any  point  in 
the  ±  is  equally  distant  from  the  extremities  of  the  line.]         (§  41,  I) 

Substituting  for  BE  its  equal  AE,  we  have 

AE  -\-  EF>  BF,  or  AF  >  BF. 

42  Cor.  I.  Every  point  which  is  equally  distant  from  the 
extremities  of  a  straight  line,  lies  in  the  perpendicular  erected 
at  the  middle  point  of  the  line. 

43.  Cor.  II.  Since  a  straight  line  is  determined  by  any 
two  of  its  points  (§  18),  it  follows  from  §  42  that 

Two  points,  each  equally  distant  from  the  extremities  of  a 
straight  line,  determine  a  perpendicular  at  its  middle  point. 

44.  Cor.  III.  When  figure  BDE  is  superposed  upon  figure 
ADE,  in  the  proof  of  §  41, 1.,  ZEBD  coincides  with  Z  EAD, 
and  Z  BED  with  Z  AED. 

That  is,  Z  EAD  =  Z  EBD,  and  Z  AED  =  Z  BED. 
Then,  if  lines  be  dratvn  to  the  extremities  of  a  straight  line 
from  any  point  in  the  perpendicular  erected  at  its  middle  pointy 

1.  They  make  equal  ayigles  with  the  line. 

2.  They  make  equal  angles  with  the  perpendicular. 


RECTILINEAR  FIGURES. 


15 


Prop.  VI.     Theorem. 

45.   From  a  given  point  without  a  straight  line,  a  perpen- 
dicular can  be  drawn  to  the  line,  and  hut  one. 

0  ^ 


H 


Given  point  C  without  line  AB. 

To  Prove  that  a  _L  can  be  drawn  from  C  to  AB,  and  but  one. 

Proof.     Let  line  HKhQ  ±  to  line  FG  at  H. 

[At  a  given  point  in  a  str.  line,  a  _L  to  the  line  can  be  drawn.]  (§  25) 

Apply  line  FO  to  line  AB,  and  move  it  along  until  HK 
passes  through  C;  let  point  //  fall  at  D,  and  draw  line  CD. 

Then,  CD  is  _L  AB. 

If  possible,  let  CE  be  another  ±  from  C  to  AB. 

Produce  CD  to  C",  making  OD  =  CD,  and  draw  line  EC. 

By  cons.,  ED  is  ±  to  CC  at  its  middle  point  D. 
.-.  ZCED  =  ZC'ED. 

[If  lines  be  drawn  to  the  extremities  of  a  str.  line  from  any  point  in 
the  ±  erected  at  its  middle  point,  they  make  equal  A  with  the  ±.] 

(§44) 

But  by  hyp.,  Z  CED  is  a  rt.  Z  ;  then,  Z  C'ED  is  a  rt.  Z. 
.-.  Z  CED  +  Z  C'ED  =  two  rt.  A. 

Then  line  CEC  is  a  str.  line. 

[If  the  sum  of  two  adj.  A  is  equal  to  two  rt.  A,  their  ext.  sides  lie 
in  the  same  str.  line.]  (§  37) 

But  this  is  impossible,  for,  by  cons.,  CDC  is  a  str.  line. 

[But  one  str.  line  can  be  drawn  between  two  points.]  (Ax,  3) 

Hence,  CE  cannot  be  _L  AB,  and  CD  is  the  only  ±  that 
can  be  drawn. 


16 


PLANE   GEOMETRY.— BOOK  I. 


Prop.  VII.     Theorem. 

46.    The  perpendicular  is  the  shortest  line  that  can  be  drawn 
from  a  point  to  a  straight  line. 


Given  CD  the  ±  from  point  C  to  line  AB,  and  CE  any 
other  str.  line  from  C  to  AB. 

To  Prove  CD  <  CE. 

Proof.  Produce  CD  to  6",  making  OD  =  CD,  and  draw 
line  EC\ 

By  cons.,  ED  is  ±  to  CC  at  its  middle  point  D. 

.'.  CE^C'E. 

[If  a  ±  be  erected  at  the  middle  point  of  a  str.  line,  any  point  in  the 
±  is  equally  distant  from  the  extremities  of  the  line.]  (§  41) 

But  CD  +  DC  <CE  +  EC. 

[A  str.  line  is  the  shortest  line  between  two  points.]  (Ax  4.) 

Substituting  for  DC  and  EC  their  equals  CD  and  CE, 
respectively,  we  have 

2CD<2  CE. 

.'.  CD<CE. 

47.  Sch.  The  distance  of  a  point  from  a  line  is  understood 
to  mean  the  length  of  the  perpendicular  from  the  point  to 
the  line. 


Ex.  9.   Find  the  number  of  degrees  in  the  angle  the  sum  of  whose 
supplement  and  complement  is  196°. 


RECTILINEAR  FIGURES.  ^7 


Prop.  VIII.    Theorem. 

48.  If  two  lines  he  drawn  from  a  point  to  the  extremities  of 
a  straight  line,  their  sum  is  greater  than  the  sum  of  two  other 
lines  similarly  drawn,  hut  enveloped  hy  them. 


Given  lines  AB  and  AC  drawn  from  point  A  to  the 
extremities  of  line  BC',  and  DB  and  DC  two  other  lines 
similarly  drawn,  but  enveloped  by  AB  and  AC. 

To  Prove  AB^AC>DB  +  DC 

Proof.    Produce  BD  to  meet  ^C  at  E. 

Now  AB->rAE>  BE. 

[A  str.  line  is  the  shortest  line  between  two  points.]  (Ax.  4) 

Adding  EC  to  both  members  of  the  inequality, 
BA  +  AC>BE  +  EC. 

Again,  DE  +  EC>DC 

Adding  BD  to  both  members  of  the  inequality, 
BE-\-EC>BD+DC. 

Since  BA  +  AC  is  greater  than  BE  -f  EC,  which  is  itself 
greater  than  BD  +  DC,  it  follows  that 

AB-\-AC>DB-]-DC. 

EXERCISES. 

10.    The  straight  line  which  bisects  an  angle 
bisects  also  its  vertical  angle. 

(If    0^  bisects    ZAOC,    ZAOE  =  ZCOE;     E 
and  these  A  are  equal  to   ABOF  and  DOF, 
respectively.) 


18 


PLANE   GEOMETRY.— BOOK  I. 


11.  The  bisectors  of  a  pair  of  vertical  angles  lie  in  the  same 
straight  line. 

(Fig.  of  Ex.  10.  To  prove  ^OFa  str.  line.  Z  COE=  ZDOF,  for 
they  are  the  halves  of  equal  A  ;  but  ZDOE  +  Z  GOE  =  2  rt.  A,  and 
therefore  ZDOE  ■\-ADOF  ='in.  A.) 

12.  The  bisectors  of  two  supplementary  ad- 
jacent angles  are  perpendicular  to  each  other. 

(We  have  ZACD  +  ZBCD  =  2  rt.  A;  and 
ADCE  and  DCF  are  the  halves  of  AACD  and 
BCD,  respectively.) 

13.  If  the  bisectors  of  two  adjacent  angles  are  perpendicular,  the 
angles  are  supplementary. 

(Fig.  of  Ex.  12.  Sum  of  ADCE  and  DCF=  1  rt.  Z,  and  ADCE 
and  DCF  are  the  halves  of  AACD  and  BCD,  respectively.) 

14.  A  line  drawn  through  the  vertex  of  an  angle     ^ 
pei*pendicular  to  its  bisector  makes  equal  angles  with 
the  sides  of  the  given  angle.  o 

(A  AOD    and    BOE  are  complements  of  AAOC 
and  BOC,  respectively.)  c 


Prop.  IX.     Theorem. 

49.  If  oblique  lines  be  drawn  from  a  point  to  a  straight 
line, 

I.    Two  oblique  lines  cutting  off  equal  distances  from  the 
foot  of  the  perpendicular  from  the  point  to  the  line  are  equal. 

II.  Of  two  oblique  lines  cutting  off  unequal  distances  from 
the  foot  of  the  perpendicular  from  the  point  to  the  line,  the 
more  remote  is  the  greater. 

C 


A    E 


F    B 


I.  Given  CD  the  ±  from  point  C  to  line  AB;  and  CE  and 
CF  oblique  lines  from  C  to  AB,  cutting  olf  equal  distances 
from  the  foot  of  CD. 


RECTILINEAR   FIGURES.  ig 

To  Prove  CE  =  CF. 

Proof.     By  hyp.,  CD  is  ±  to  EF  at  its  middle  point  D. 

.-.  CE=CF. 

[If  a  _L  be  erected  at  the  middle  point  of  a  str.  line,  any  point  in 
the  ±  is  equally  distant  from  the  extremities  of  the  line.]  (§  41) 


II.  Given  CD  the  X  from  point  C  to  line  AB;  and  CE 
and  CF  oblique  lines  from  C  to  AB,  cutting  off  unequal 
distances  from  the  foot  of  CD ;  CF  being  the  more  remote. 

To  Prove  CF>CE. 

Proof.  Produce  CD  to  C",  making  CD  =  CD,  and  draw 
lines  CE  and  CF. 

By  cons.,  AD  is  ±  to  CC"  at  its  middle  point  D. 

.-.  CF=  CF,  and  CE  =  CE. 

[If  a  ±  be  erected  at  the  middle  point  of  a  str.  line,  any  point  in 
the  ±  is  equally  distant  from  the  extremities  of  the  line.]  (§  41) 

But  CF  +  FC  >CE  +  EC. 

[If  two  lines  be  drawn  from  a  point  to  the  extremities  of  a  str.  line, 
their  sum  is  <  the  sum  of  two  other  lines  similarly  drawn,  but  en- 
veloped by  them.]  (§  48) 

Substituting  for  FO  and  EC  their  equals  CF  and  CE, 

respectively,  we  have 

2CF>2CE. 

.'.  CF>CE. 

Note.  The  theorem  holds  equally  if  oblique  line  CE  is  on  the 
opposite  side  of  perpendicular  CD  from  CF. 


20  PLANE  GEOMETRY.— BOOK  I. 

Prop.  X.    Theorem. 

50.  (Converse  of  Prop.  IX.,  I.)  If  oblique  lines  he  drawn 
from  a  point  to  a  straight  liiie,  two  equal  oblique  lines  cut  off 
equal  distances  from  the  foot  of  the  perpendicular  from  the 
point  to  the  line. 

G 


A    E  D  F    B 


Given  CD  the  ±  from  point  C  to  line  AB,  and  CE  and 
CF  equal  oblique  lines  from  C  to  AB. 

To  Prove  DJE  =  DF. 

Proof.  We  know  that  DE  is  either  >,  equal  to,  or  < 
DF. 

If  we  suppose  DE  >  DF,  CE  would  be  >  CF. 

[If  oblique  lines  be  drawn  from  a  point  to  a  str.  line,  of  two  oblique 
lines  cutting  off  unequal  distances  from  the  foot  of  the  ±  from  the 
point  to  the  line,  the  more  remote  is  the  greater.]  (§  49) 

But  this  is  contrary  to  the  hypothesis  that  CE  =  CF. 

Hence,  DE  cannot  be  >  DF. 

In  like  manner,  if  we  suppose  DE  <  DF,  CE  would  be 
<  CF,  which  is  contrary  to  the  hypothesis  that  CE  =  CF. 

Hence,  DE  cannot  be  <  DF. 

Then,  if  DE  can  be  neither  >  DF,  nor  <  DF,  we  must 
have  DE=DF. 

Note.  The  method  of  proof  exemplified  in  Prop.  X  is  known  as 
the  "  Indirect  Method,"  or  the  "  Beductio  ad  Ahsurdum.'''' 

The  truth  of  a  proposition  is  demonstrated  by  making  every  pos- 
sible supposition  in  regard  to  the  matter,  and  sliowing  that,  in  all 
cases  except  the  one  which  we  wish  to  prove,  the  supposition  leads  to 
something  which  is  contrary  to  the  hypothesis. 


RECTILINEAR  FIGURES.  21 

51.  Cor.  (Converse  of  Prop.  IX,  II.)  If  two  unequal 
oblique  lines  he  drawn  from  a  point  to  a  straight  line,  the 
greater  cuts  off  the  greater  distance  from 

the  foot  of  the  perpendicular  from  the 
point  to  the  line. 

Given  CD  the  _L  from  point  C  to  line 
AB ;  and  GE  and  CF  unequal  oblique 
lines  from  C  to  AB,  CF  being  >  CE. 

To  Prove  DF>DE. 

(Prove  by  Reductio  ad  Absurdum;  by  §  49,  I,  DE  cannot 
equal  DF,  and  by  §  49,  II,  it  cannot  be  >  DF?) 

PARALLEL  LINES. 

52.  Def.     Two  straight  lines  are  said  to  be  parallel  (II) 

when  they  lie  in  the  same  plane,  and    ^ ^ 

cannot  meet  however  far  they  may  be 

produced;  as  ^5  and  CD.  ^  ^ 

53.  Ax.  We  assume  that  hut  one  straight  line  can  he 
drawn  through  a  given  point  parallel  to  a  given  straight  line. 

Prop.  XI.     Theorem. 

54.  Two  perpendiculars  to  the  same  straight  line  are 
parallel.  . 


Given  lines  AB  and  CD  A.  to  line  AC. 

To  Prove  AB  II  CD. 

Proof.  If  AB  and  CD  are  not  II,  they  will  meet  in  some 
point  if  sufficiently  produced  (§  52). 

We  should  then  have  two  Js  from  this  point  to  AC,  which 
is  impossible. 

[From  a  given  point  without  a  str.  line,  but  one  ±  can  be  drawn  to 
the  line.]  (§  ^5) 

Therefore,  AB  and  CD  cannot  meet,  and  are  II. 


22  PLANE   GEOMETRY.  —  BOOK  I. 


Prop.  XII.     Theorem. 

55.    Two  straight  lines  parallel  to  the  same  straight  line  are 
parallel  to  each  other. 

A B 


E F 

Given  lines  AB  and  CD  II  to  line  EF. 

To  Prove  AB  II  CD. 

Proof.  If  AB  and  CD  are  not  II,  they  will  meet  in  some 
point  if  sufficiently  produced.  (§  52) 

We  should  then  have  two  lines  drawn  through  this  point 
II  to  EF,  which  is  impossible. 

[But  one  str.  line  can  be  drawn  through  a  given  point  II  to  a  given 
str.  line.]  (§  53) 

Therefore,  AB  and  CD  cannot  meet,  and  are  II. 

Prop.  XIII.     Theorem. 

56.  A  straight  line  perpendicular  to  one  of  two  parallels  is 
perpendicular  to  the  other. 


Given  lines  AB  and  CD  II,  and  line  AC  ±  AB. 
To  Prove  AC  A.  CD. 

Proof.     If  CD  is  not  ±  AC,  let  line  CE  be  ±  AC. 
Then  since  AB  and  CE  are  ±  AC,  CE  II  AB. 
[Two  Js  to  the  same  str.  line  are  II.]  (§  54) 

But  by  hyp.,  CD  II  AB. 

Then,  CE  must  coincide  with  CD. 

[But  one  str.  line  can  be  drawn  through  a  given  point  II  to  a  given 
str.  line.]  (§53) 


RECTILINEAR  FIGURES.  23 

But  by  hyp.,  AC  ±  CE. 

Then  since  CE  coincides  with  OD,  we  have  AC  ±  CD. 

TRIANGLES. 
DEFINITIONS. 

57.  A  triangle  (A)  is  a  portion  of  a  plane  bounded  by 
three  straight  lines ;  as  ABC. 

The  bounding  lines,  AB,  BC,  and  CA, 
are  called  the  sides  of  the  triangle,  and 
their  points  of  intersection.  A,  B,  and  (7, 
the  vertices. 

The  angles    of    the   triangle   are  the 
angles  GAB,  ABC,  and  BCA,  included  between  the  adjacent 
sides. 

An  exterior  angle  of  a  triangle  is 
the  angle  at  any  vertex  between  any 
side  of  the  triangle  and  the  adjacent 
side  produced ;  as  ACD. 

58.  A  triangle  is  called  scalene  when  no  two  of  its  sides 
are  equal ;  isosceles  when  two  of  its  sides  are  equal ;  equi- 
lateral when  all  its  sides  are  equal ;  and  equiangular  when 
all  its  angles  are  equal. 


Scalene.  Isosceles.  Equilateral. 

59.  A  right  triangle  is   a  triangle  which   has   a  right 
angle;     as  ABC,   which   has    a   right  ^A 

angle  at  C. 

The  side  AB  opposite  the  right  angle 
is  called  the  hypotenuse,  and  the  other 
sides,  AC  and  BC,  the  legs. 


24 


PLANE   GEOMETRY.— BOOK  I. 


60.  If  any  side  of  a  triangle  be  taken  and  called  the 
base,  the  corresponding  altitude  is  the  perpendicular  drawn 
from  the  opposite  vertex  to  the  base,  produced  if  necessary. 

In  general,  either  side  may  be  taken  as  the  base ;  but  in 
an  isosceles  triangle,  unless  otherwise  specified,  the  side 
which  is  not  one  of  the  equal  sides  is  taken  as  the  base. 

When  any  side  has  been  taken  as  the 
base,  the  opposite  angle  is  called  the  ve?-- 
tical  angle,  and  its  vertex  is  called  the 
vertex  of  the  triangle. 

Thus,  in  triangle  ABC,  BC  is  the  base, 
AD  the  altitude,  and  BAG  the  vertical 
angle. 

61.  Since  a  straight  line  is  the  shortest  line  between  two 
points  (Ax.  4),  it  follows  that 

Any  side  of  a  triangle  is  less  than  the  sum  of  the  other 
two  sides. 


Prop.  XIV.     Theorem. 

62.  Any  side  of  a  triangle  is  greater  than  the  difference  of 
the  other  two  sides. 


Given  AB,  any  side  of  A  ABC;  and  side  ^C>  side  AC. 
To  Prove  AB>BC-  AC. 

Proof.     We  have  AB-}-AC>BC. 

[A  str.  line  is  the  shortest  line  between  two  points.]  (Ax.  4) 

Subtracting  AC  from  both  members  of  the  inequality, 
AB>BC-Aa 


RECTILINEAR  FIGURES.  25 

Prop.  XV.    Theorem. 

63.  Ti(}o  triangles  are  equal  when  two  sides  and  the  in- 
cluded angle  of  one  are  equal  respectively  to  two  sides  and  the 
included  angle  of  the  other. 


A  B  D 

Given,  in  A  ABC  and  DEF, 

AB  =  DE,  AC=DF,  and  ZA  =  ZD. 

To  Prove  A  ABC  =  A  DEF. 

Proof.  Superpose  A  ABC  upon  A  DEF  in  such  a  way 
that  Z  A  shall  coincide  with  its  equal  Z  D ;  side  AB  falling 
on  side  DE,  and  side  AC  on  side  DF. 

Then  since  AB  =  DE  and  AC  =  DF,  point  B  will  fall  on 
point  E,  and  point  C  on  point  F. 

Whence,  side  BC  will  coincide  with  side  EF. 

[But  one  str.  line  can  be  drawn  between  two  points.]  (Ax.  3) 

Therefore,  the  A  coincide  throughout,  and  are  equal. 

64.  Cor.  Since  ABC  and  DEF  coincide  throughout,  we 
have  ZB  =  ZE,  ZC=ZF,  8^nd  BC  =  EF. 

65.  Sch.  I.  In  equal  figures,  lines  or  angles  which  are 
similarly  placed  are  called  homologous. 

Thus,  in  the  figure  of  Prop.  XV,  Z  A  is  homologous  to 
Z  D ;  AB  is  homologous  to  DE ;  etc. 

66.  Sch.  II.     It  follows  from  §  65  that 

In  equal  figures,  the  homologous  parts  are  equal. 

67.  Sch.  III.  In  equal  triangles,  the  equal  angles  lie 
opposite  the  equal  sides. 


26  PLANE   GEOMETRY.— BOOK  I. 

Pkop.  XVI.    Theorem. 

68.  Two  triangles  are  equal  when  a  side  and  two  adjacent 
angles  of  one  are  equal  respectively  to  a  side  and  two  adjaxient 
angles  of  the  other. 

9.  F 


Given,  in  A  ABC  and  DEF, 

AB  =  DE,  ZA  =  ZD,  and  ZB  =  ZE. 
To  Prove  A  ABC  =  A  DEF. 

Proof.  Superpose  A  ABC  upon  A  DEF  in  such  a  way 
that  side  AB  shall  coincide  with  its  equal  DE-,  point  A 
falling  on  point  D,  and  point  B  on  point  E. 

Then  since  ZA  =  ZD,  side  AC  will  fall  on  side  DF,  and 
point  C  will  fall  somewhere  on  DF. 

And  since  ZB  =  Z  E,  side  BC  will  fall  on  side  EF,  and 
point  C  will  fall  somewhere  on  EF. 

Then  point  C,  falling  at  the  same  time  on  DF  and  EF, 
must  fall  at  their  intersection,  F. 

Therefore,  the  A  coincide  throughout,  and  are  equal. 

EXERCISES. 

15.  If,  in  the  figure  of  Prop.  XV.,  AB=EF,  BC=DE,  and 
ZB  =  ZE,  whicli  angle  of  triangle  DEF  is  equal  to  J.  ?  which  angle 
is  equal  to  C  ? 

16.  K,  in  the  figure  of  Prop.  XVI.,  AC  =  DF,  ZA  =  ZF,  and 
ZC=  ZD,  which  side  of  triangle  DEF  is  equal  to  AB ?  which  side 
is  equal  to  BC? 

17.  If  OD  and  OE  are  the  bisectors  of  two  complementary- 
adjacent  angles,  AOB  and  BOC,  how  many  degrees  are  there  in 
ZDOE? 


RECTILINEAR  FIGURES. 


27 


Prop.  XVII.     Theorem. 

69.    Two  triangles  are  equal  when  the  three  sides  of  one  are 
equal  respectively  to  the  three  sides  of  the  other. 


Given,  in  A  ABC  and  DEF, 

AB  =  BE,  BC  =  EF,  and  CA  =  FD. 

To  Prove  A  ABC  =  A  DEF. 

Proof.  Place  A  DEF  in  the  position  ABF;  side  DE 
coinciding  with  its  equal  AB,  and  vertex  F  falling  at  F,  on 
the  opposite  side  of  AB  from  C. 

Draw  line  CF\ 

By  hyp.,  AC  =  AF^  and  BC  =  BF'. 

Whence,  AB  is  _L  to  CF'  at  its  middle  point. 

[Two  points,  each  equally  distant  from  the  extremities  of  a  str.  line, 
determine  a  ±  at  its  middle  point.  ]  (§  43) 

.-.  ZBAC=ZBAF'. 

[If  lines  be  drawn  to  the  extremities  of  a  str.  line  from  any  point 
in  the  ±  erected  at  its  middle  point,  they  make  equal  A  with  the  ±.] 

(§44) 

Then  since  sides  AB  and  AC  and  Z  BAC  of  A  ABC  are 

equal,  respectively,  to  sides  AB  and  AF'  and  Z  BAF'  of 

AABF', 

AABC=AABF'. 

[Two  A  are  equal  when  two  sides  and  the  included  Z  of  one  are 
equal  respectively  to  two  sides  and  the  included  Z  of  the  other.] 

(§63) 
That  is,  A  ABC  =  A  DEF. 


28 


PLANE   GEOMETRY. —BOOK  I. 


Prop.  XVIII.     Theorem. 

70.  Two  right  triangles  are  equal  when  the  hypotenuse  and 
an  adjacent  angle  of  one  are  equal  respectively  to  the  hypote- 
nuse and  an  adjacent  angle  of  the  other. 

E 


A  CD 

Given,  in  rt.  A  ABC  and  DEF, 

hypotenuse  AB  =  hypotenuse  BE,  and  Z  A  =  Z.D. 
To  Prove  A  ABC  =  A  DEF. 

Proof.  Superpose  A  ABC  upon  A  DEF  in  such  a  way 
that  hypotenuse  AB  shall  coincide  with  its  equal  DE ;  point 
A  falling  on  point  D,  and  point  B  on  point  E. 

Then  since  Z.A  =  ZD,  side  AC  will  fall  on  side  DF. 

Therefore,  side  BC  will  fall  on  side  EF. 

[From  a  given  point  without  a  str.  line,  but  one  _L  can  be  drawn  to 
the  line.]  (§45) 

Therefore,  the  A  coincide  throughout,  and  are  equal. 

71.  Def .  If  two  straight  lines,  AB 
and  CD,  are  cut  by  a  line  EF,  called 
a  transversal,  the  angles  are  named 
as  follows : 

c,  d,  e,  and  /  are   called   interior 


and   a,  b. 


and  h   exterior 


angles, 
angles. 

c  and  /,  or  d  and  e,  are  called  alter- 
nate-interior angles. 

a  and  h,  or  h  and  g,  are  called  alternate-exterior  angles. 

a  and  e,  b  and  /,  c  and  g,  or  d  and  h,  are  called  corre- 
sponding angles. 


RECTILINEAR  FIGURES.  29 


Prop.  XIX.     Theorem. 


72.   If  two  parallels  are  cut  by  a  transversal^  the  alternate- 
interior  angles  are  equal. 


-B 


Given  lis  AB  and  CD  cut  by  transversal  EF  at  points  G 
and  H,  respectively. 

To  Prove  Z  AGH=  Z  OHD  and  Z  BGH=Z  OHG. 

Proof.     Through  K,  the  middle  point  of  GH,  draw  line 
LMl.  AB',  then,  LM 1.  CD. 

[A  str.  line  _L  to  one  of  two  ||s  is  ±  to  the  other.]  (§  56) 

Now  in  rt.  A  GKL  and  HKM,  by  cons., 

hypotenuse  GK  =  hypotenuse  HK. 

Also,  Z  GKL  =  Z  HKM. 

[If  two  str.  lines  intersect,  the  vertical  A  are  equal.]  (§  40) 

.-.  A  GKL  =  A  HKM. 

[Two  rt.  ^  are  equal  when  the  hypotenuse  and  an  adj.  Z  of  one  are 
equal  respectively  to  the  hypotenuse  and  an  adj.  Z  of  the  other.]  (§  70) 

.-.  ZKGL  =  ZKHM. 

[In  equal  figures,  the  homologous  parts  are  equal.]  (§  QQ) 

Again,  Z  KGL  is  the  supplement  of  Z  BGH,  and  Z  KHM 

the  supplement  of  Z  CHG. 

[If  two  adj.  A  have  their  ext.  sides  in  the  same  str.  line,  they  are 

supplementary.]  (§  33) 

Then  since  Z  KGL  =  Z  KHM,  we  have 
ZBGH=ZCHG, 

[The  supplements  of  equal  A  are  equal]  (§  31) 


30  PLANE   GEOMETRY.— BOOK  I. 


Prop.  XX.     Theorem. 

73.  (Converse  of  Prop.  XIX.)  If  two  straight  lines  are 
cut  by  a  transversal,  and  the  alternate-interior  angles  are  equal, 
the  tivo  lines  are  parallel. 


Given  lines  AB  and  CD  cut  by  transversal  EF  at  points 
G  and  H,  respectively,  and 

Z  AGH=  Z  GHD. 

To  Prove  AB  II  CD. 

Proof.    If  CD  is  not  II  AB,  draw  line  KL  through  H II  AB. 
Then  since  lis  AB  and  KL  are  cut  by  transversal  EF, 

ZAGH=ZGHL. 

[If  two  lis  are  cut  by  a  transversal,  the  alt.  int.  A  are  equal.]    (§  72) 

But  by  hyp.,  Z  AGH  =  Z  GHD. 

.'.  ZGHL  =  ZGHD. 

[Things  which  are  equal  to  the  same  thing  are  equal  to  each  other.] 

(Ax.  1) 
But  this  is  impossible  unless  KL  coincides  with  CD. 

.:  CDWAB. 

In  like  manner,  it  may  be  proved  that  if  AB  and  CD  are 
cut  by  EF,  and  Z  BGH=  Z  CHG,  then  AB  II  CD. 


Ex.  18.   If,  in  the  figure  of  Prop.  XIX.,  /LAGH=Q^°,  how  many 
degrees  are  there  in  BGW}  in  GHD  ?  in  DHF'i 


RECTILINEAR  FIGURES.  3J^ 

Prop.  XXI.    Theorem. 

74.  If  two  parallels  are  cut  by  a  transversal,  the  correspond- 
ing angles  are  equal. 


Given  lis  AB  and  CD  cut  by  transversal  EF  at  points  G 
and  H,  respectively. 

To  Prove  Z  AGE  =  Z  CHG. 

Proof.     We  have    ZBGH=ZCHG. 

[If  two  lis  are  cut  by  a  transversal,  the  alt.  int.  A  are  equal.]     (§  72) 

But,  ZBGH=ZAGE. 

[If  two  str.  lines  intersect,  the  vertical  A  are  equal.]  (§  40) 

.-.  ZAGE  =  ZCHG. 

[Things  which  are  equal  to  the  same  thing  are  equal  to  each 
other.]  (Ax.  1) 

In  like  manner,  we  may  prove 

Z  AGH=Z  CHF,  Z  BGE=Z  DHG,  and  Z  BGH=Z DHF. 

75.  Cor.  I.  If  two  parallels  are  cut  by  a  transversal,  the 
alternate-exterior  angles  are  equal. 

(Fig.  of  Prop.  XXI.) 

Given  lis  AB  and  CD  cut  by  transversal  EF  at  points  G 
and  H,  respectively. 

To  Prove  Z  AGE  =  Z  DHF. 

(Z  BGH=  Z  CHG,  and  the  theorem  follows  by  §  40.) 

What  other  two  ext.  A  in  the  figure  are  equal  ? 


32  PLANE   GEOMETRY.— BOOK  I. 

76.  Cor.  II.  If  two  parallels  are  cut  by  a  transversal,  the 
sum  of  the  interior  angles  on  the  same  side  of  the  transversal 
is  equal  to  two  right  angles. 

(Fig.  of  Prop.  XXI.) 

Given  lis  AB  and  CD  cut  by  transversal  EF  at  points  G 
and  H,  respectively. 

To  Prove      Z  AGH  +  Z  CHG  =  two  rt.  A. 

(By  §  32,  /.AGH+ZAGE  =  t^o  rt.  A-,  the  theorem 
follows  by  §  74.) 

What  other  two  int.  A  in  the  figure  have  their  sum  equal 
to  two  rt.  A  ? 

Prop.  XXII.     Theorem. 

77.  (Converse  of  Prop.  XXI.)  If  two  straight  lines  are 
cut  by  a  transversal,  and  the  corresponding  angles  are  equal, 
the  two  lines  are  parallel. 


'E 


Given  lines  AB  and  CD  cut  by  transversal  EF  at  points 
G  and  H,  respectively,  and 

Z  AGE  =  Z  CHG. 
To  Prove  AB  II  CD. 

Proof.     We  have  Z  AGE  =  Z  BGH.  -     , 

[If  two  str.  lines  intersect,  the  vertical  A  are  equal.]  (§  40) 

.-.  ZBGH=^ZCHG. 

[Things  which  are  equal  to  the  same  thing  are  equal  to  each  other.] 

(Ax.  1) 


RECTILINEAR  FIGURES.  33 

.-.  ABWCD. 

[If  two  str.  lines  are  cut  by  a  transversal,  and  the  alt.  int.  A  are 
equal,  the  two  lines  are  II.]  (-§  73) 

In  like  manner,  it  may  be  proved  that  if 

Z  AGH=Z  CIIF,  or  Z  BGE=ZDHG,  or  Z  BGH=ZDHF, 

then  AB  11  CD. 

78.  Cor.  I.  (Converse  of  §  75.)  If  two  straight  lines 
are  cut  by  a  transversal,  and  the  alternate-exterior  angles  are 
equal,  the  two  lines  are  parallel. 

(Fig.  of  Prop.  XXII.) 

Given  lines  AB  and  CD  cut  by  transversal  EF  at  points 
G  and  H,  respectively,  and 

Z  AGE  =  Z  DHF. 

To  Prove  AB  11  CD. 

(Z  AGE  =  Z  BGH,  and  Z  DHF=  Z  CEG;  and  the  theo- 
rem follows  by  §  73.) 

What  other  two  ext.  A  are  there  in  the  figure  such  that, 
if  they  are  equal,  AB  II  CD  ? 

79.  Cor.  II.  (Converse  of  §  76.)  If  two  straight  lines 
are  cut  by  a  transversal,  and  the  sum  of  the  interior  angles  on 
the  same  side  of  the  transversal  is  equal  to  two  right  angles, 
the  two  lines  are  parallel. 

(Fig.  of  Prop.  XXII.) 

Given  lines  AB  and  CD  cut  by  transversal  EF  at  points 
G  and  H,  respectively,  and 

Z  AGH+  Z  CHG  =  two  rt.  A. 

To  Prove  AB  II  CD. 

(Z  CHG  is  the  supplement  of  Z  AGH,  and  also  of  Z  GHD ; 
then  A  AGH  and  GHD  are  equal  by  §  31,  2,  and  the  theo- 
rem follows  by  §  73.) 

What  other  two  int.  A  are  there  in  the  figure  such  that,  if 
their  sum  equals  two  rt.  A,  AB  II  CD  ? 


34  PLANE   GEOMETRY.— BOOK  I. 

Prop.  XXIII.     Theorem. 
80.    Two  parallel  lines  are  everywhere  equally  distant. 
A     E F    B 


C       G  H     D 


Given  lis  AB  and  CD,  E  and  F  any  two  points  on  AB, 
and  Ea  and  FH  lines  ±  CD. 

To  Prove  EG  =  FH  (§  47). 

Proof.     Draw  line  FG. 

We  have  EG±AB. 

[A  str.  line  i.  to  one  of  two  ||s  is  ±  to  the  other.]  (§  56) 

Then,  in  rt.  A  EFG  and  FGH, 

FG  =  FG. 

And  since  lis  AB  and  CD  are  cut  by  FG, 

Z  EFG  =  Z  FGH. 

[If  two  lis  are  cut  by  a  transversal,  the  alt.  int.  A  are  equal.]     (§  72) 

.-.  A  EFG  =  A  FGH. 

[Two  rt.  A  are  equal  when  the  hypotenuse  and  an  adj.  Z  of  one  are 
equal  respectively  to  the  hypotenuse  and  an  adj.  Z  of  the  other.]    (§  70) 

.-.  EG  =  FH 

[In  equal  figures,  the  homologous  parts  are  equal.]  (§  66) 

Prop.  XXIV.     Theorem. 

81.  Two  angles  whose  sides  are  parallel,  each  to  each,  are 
equal  if  both  pairs  of  parallel  sides  extend  in  the  same  direc- 
tion, or  171  opposite  directions,  from  their  vertices. 

Note.  The  sides  extend  in  the  same  direction  if  they  are  on  the 
same  side  of  a  straight  line  joining  the  vertices,  and  in  opposite  direc- 
tions if  they  are  on  opposite  sides  of  this  line. 


RECTILINEAR  FIGURES.  35 

fA        /D 


Given  lines  AB  and  BC  \\  to  lines  DH  and  KF,  respec- 
tively, intersecting  at  E. 

I.  To  Prove  that  A  ABC  and  DEF,  whose  sides  AB  and 
DE,  and  also  BC  and  ^i^,  extend  in  the  same  direction 
from  their  vertices,  are  equal. 

Proof.     Let  BC  and  DH  intersect  at  G. 
Since  lis  AB  and  DE  are  cut  by  BC, 

ZABC=ZDGC 

[If  two  lis  are  cut  by  a  transversal,  the  corresp.  A  are  equal.] 

(§74) 
In  like  manner,  since  lis  BC  and  EF  are  cut  by  DE, 

ZDGC  =  ZDEF. 

.-.  ZABG  =  ZDEF.  (1) 

[Things  which  are  equal  to  the  same  thing  are  equal  to  each  other.] 

(Ax.  1) 

II.  To  Prove  that  A  ABC  and  HEK,  whose  sides  AB 
and  EH,  and  also  BC  and  EK,  extend  in  opposite  directions 
from  their  vertices,  are  equal. 

Proof.   From(l),   Z  ABC  =  Z  DEF. 

But,  Z  DEF  =  Z  HEK. 

[If  two  str.  lines  intersect,  the  vertical  A  are  equal.]  (§  40) 

.-.  ZABC  =  ZHEK. 

[Things  which  are  equal  to  the  same  thing,  are  equal  to  each  other.] 

(Ax.  1) 


36 


PLANE   GEOMETRY.— BOOK  I. 


82.   Cor.     Two  angles  whose  sides  are  parallel,  each  to  each, 
are  supplementary  if  one  pair  of  parallel 
sides  extend   in  the  same  direction,  and 
the  other  pair  in  opposite  directions,  from 
their  vertices. 

Given  lines  AB  and  BC  \\  to  lines 
DH  and  KF,  respectively,  intersecting 
at  J57. 

To  Prove  that  A  ABC  and  DEK, 
whose  sides  AB  and  DE  extend  in  the  same  direction,  and 
BC  and  EK  in  opposite  directions,  from  their  vertices,  are 
supplementary. 

Proof.   We  have    Z  ABC  =  Z  DEF. 

[Two  A  whose  sides  are  ||,  each  to  each,  are  equal  if  both  pairs  of  || 
sides  extend  in  the  same  direction  from  tlieir  vertices.]  (§  81) 

But  Z  DEF  is  the  supplement  of  Z  DEK. 
[If  two  adj.  A  have  tlieir  ext.  sides  in  the  same  str.  line,  they  are 
supplementary.]  (§  33) 

Then  its  equal,  Z  ABC,  is  the  supplement  of  Z  DEK. 


Prop.  XXV.     Theorem. 

83.    Two  angles  whose  sides  are  perpejidicular,  each  to  each, 
are  either  equal  or  supplementary. 


Given  lines  AB  and  BC  ±  to  lines  DE  and  FG,  respec- 
tively, intersecting  at  E. 

To  Prove  Z  ABC  equal  to  Z  DEF,  and  supplementary  to 
Z  DEC. 


RECTILINEAR   FIGURES.  37 

Proof.     Draw  line  EH±  DE,  and  line  EKl.  EF. 
Then  since  EH  and  AB  are  ±  DE, 

EH  II  AB. 
[Two  ±s  to  the  same  str,  line  are  II.]  (§  54) 

In  like  manner,  since  EK  and  BC  are  _L  EF, 
EK II  BO. 
.'.  /.HEK=ZABC. 
[Two  A  whose  sides  are  ||,  each  to  each,  are  equal  if  both  pairs  of  || 
sides  extend  in  the  same  direction  from  their  vertices.]  (§  81) 

But  since,  by  cons.,  A  DEH  and  FEK  are  rfc.  A,  each  of 
the  A  DEF  and  HEK  is  the  complement  of  Z  FEH. 
.'.  A  DEF  =  A  HEK. 
[The  complements  of  equal  A  are  equal.]  (§  31) 

.-.  A  ABC  =  A  DEF. 
[Things  which  are  equal  to  the  same  thing  are  equal  to  each  other.] 

(Ax.  1) 
Again,  A  DEF  is  the  supplement  of  A  DEG. 
[If  two  adj.  A  have  their  ext.  sides  in  the  same  str.  line,  they  are 
supplementary.]  (§33) 

Then,  its  equal,  A  ABC  is  the  supplement  of  A  DEG. 

Note.  The  angles  are  equal  if  they  are  both  acute  or  both  obtuse  ; 
and  supplementary  if  one  is  acute  and  the  other  obtuse. 

EXERCISES. 

19.  If,  in  the  figure  of  Prop.  XXIV.,  Z^BC'=59°,  how  many 
degrees  are  there  in  each  of  the  angles  formed  about  the  point  E  ? 

20.  The  line  passing  through  the  vertex  of  an  angle  perpendicular 
to  its  bisector  bisects  the  supplementary  adjacent  angle. 

(Fig.  of  Ex.  12.  Let  CE  bisect  ZACD,  and  suppose  CFl.  CE ; 
sum  of  AACD  and  BCD  =  2  rt.  A;  then  sum  of  ADCE  and 
^  J?CZ)  =  1  rt.  Z  ;  but  sum  ot  ADCE  and  DCF  is  also  1  rt.  Z;  whence 
the  theorem  follows.) 

21.  Any  side  of  a  triangle  is  less  than  the  half-sum  of  the  sides  of 
the  triangle. 

(Fig.  of  Prop.  XIV.  We  have  AB<BC-^  CA;  then  add  AB  to 
both  members  of  the  inequality.) 


38  PLANE   GEOMETRY.— BOOK  I. 


Prop.  XXVI.     Theorem. 

84.    Tlie  sum  of  the  angles  of  any  triangle  is  equal  to  two 
right  angles. 


A  CD 

Given  A  ABC. 

To  Prove      Z.  A  +  /.  B  +  AC  =t^oit.  A. 
Proof.     Produce  AC  to  D,  and  draw  line  GE  II  AB. 
Then,  Z  BCD  +  Z  BCE  +  Z  AGB  =  two  rt.  A.  (1) 

[The  sum  of  all  the  A  on  the  same  side  of  a  str.  line  at  a  given 
point  is  equal  to  two  rt.  A.']  (§  34) 

Now  since  lis  AB  and  CE  are  cut  by  AD, 

A  ECI)  =  AA. 
[If  two  lis  are  cut  by  a  transversal,  the  corresp.  A  are  equal.]    (§  74) 
And  since  lis  AB  and  CE  are  cut  by  BC, 

A  BCE  =  AB. 
[If  two  lis  are  cut  by  a  transversal,  the  alt.  int.  A  are  equal.]    (§  72) 
Substituting  in  (1),  we  have 

AA  +  AB  +  A  ACB  =  two  rt.  A. 

85.  Cor.  I.    It  follows  from  the  above  demonstration  that 
Z  BCD  =  A  ECD  +  Z  BCE  =  AA  +  AB',  hence 

1.  An  exterior  angle  of  a  triangle  is  equal  to  the  sum  of  the 
two  opposite  interior  angles. 

2.  An  exterior  angle  of  a  triangle  is  greater  than  either  of 
the  opposite  interior  angles. 

86.  Cor.  II.  If  two  triangles  have  two  angles  of  one  equal 
respectively  to  two  angles  of  the  other,  the  third  angle  of  the 
first  is  equal  to  the  third  angle  of  the  second. 


RECTILINEAR  FIGURES. 


39 


87.  Cor.  III.  A  triangle  cannot  have  two  right  angles,  nor 
two  obtuse  angles. 

88.  Cor.  rV.  The  sum  of  the  acute  angles  of  a  right  tri- 
angle is  equal  to  one  right  angle. 

89.  Cor.  V.  Two  right  triangles  are  equal  when  a  leg  and 
an  acute  angle  of  one  are  equal  respectively  to  a  leg  and  the 
homologous  acute  angle  of  the  other. 

The  theorem  follows  by  §§  86  and  68. 

Prop.  XXVII.     Theorem. 

90.  Two  right  triangles  are  equal  when  the  hypotenuse  and 
a  leg  of  one  are  equal  respectively  to  the  hypotenuse  and  a  leg 
of  the  other. 


A  CD 

Given,  in  rt.  A  ABC  and  DEF, 

hypotenuse  AB  =  hypotenuse  DE,  and  BG  =  EF. 

To  Prove  A  ABC  =  A  DEF. 

Proof.  Superpose  A  ABC  upon  A  DEF  in  such  a  way 
that  side  BC  shall  coincide  with  its  equal  EF;  point  B 
falling  on  point  E,  and  point  C  on  point  F. 

We  have  ZC=ZF. 

[All  rt.^  are  equal.]  (§26) 

Then,  side  AC  will  fall  on  side  DF. 

But  the  equal  oblique  lines  AB  and  DE  cut  ofP  upon  DF 
equal  distances  from  the  foot  of  J_  EF. 

[If  oblique  lines  be  drawn  from  a  point  to  a  str.  line,  two  equal 
oblique  lines  cut  off  equal  distances  from  the  foot  of  the  ±  from  the 
point  to  the  line.]  (§  ^0) 

Therefore,  point  A  falls  on  point  D. 

Hence,  the  A  coincide  throughout,  and  are  equal. 


40  PLANE   GEOMETRY.  — BOOK  I. 

Prop.  XXVIII.     Theorem. 

91.  If  two  triangles  have  two  sides  of  one  equal  respectively 
to  two  sides  of  the  other,  but  the  included  angle  of  the  first 
greater  than  the  included  angle  of  the  second,  the  third  side 
of  the  first  is  greater  than  the  third  side  of  the  second. 


E 

a 

Given,  in  A  ABC  and  DEF, 

AB  =  DE,  AO=DF,  and  ZBAC>ZD. 
To  Prove  BC>EF. 

Proof.  Place  A  DEF  in  the  position  ABO;  side  DE 
coinciding  with  its  equal  AB,  and  vertex  F  falling  at  G. 

Draw  line  AH  bisecting  Z  GAG,  and  meeting  BG  at  H; 
also,  draw  line  GH. 

In  A  AGH  and  AGH,  AH=  AH. 

Also,  by  hyp.,  AG  =  AG. 

And  by  cons.,  Z  GAH=  A  GAH. 

.:  A  AGH  =  A  AGH 

[Two  ^  are  equal  when  two  sides  and  the  included  Z  of  one  are  equal 
respectively  to  two  sides  and  the  included  Z  of  the  other.]  (§  63) 

.-.  GH=GH. 

[In  equal  figures,  the  homologous  parts  are  equal.]  (§  66) 

But,  BH+GH>BG. 

[A  str.  line  is  the  shortest  line  between  two  points.]  (Ax.  4) 

Substituting  for  GH  its  equal  GH,  we  have 

BH+GH>BG,  or  BG>EF. 


RECTILINEAR  FIGURES. 


41 


Prop.  XXIX.     Theorem. 

92.  (Converse  of  Prop.  XXVIII.)  //  two  triangles  have 
two  sides  of  one  equal  respectively  to  two  sides  of  the  other, 
but  the  third  side  of  the  first  greater  than  the  third  side  of  the 
second,  the  included  angle  of  the  first  is  greater  than  the 
included  angle  of  the  second. 


Given,  in  A  ABC  and  DBF, 

AB  =  DE,  AG=DF,  and  BC>EF. 

To  Prove  /.A>^D. 

Proof.    We  know  that  ZAis  either  <,  equal  to,  or  >  Z  Z>. 

If  we  suppose  ZA  =  ZD,  A  ABC  would  equal  A  DEF. 

[Two  A  are  equal  when  two  sides  and  the  included  Z  of  one  are 
equal  respectively  to  two  sides  and  the  included  Z  of  the  other.]  (§  63) 

Then,  BC  would  equal  EF. 

[In  equal  figures,  the  homologous  parts  are  equal.]  (§  66) 

Again,  if  we  suppose  ZA<ZD,  BC  would  be  <  EF. 

[If  two  i^  have  two  sides  of  one  equal  respectively  to  two  sides  of 
the  other,  but  the  included  Z  of  the  first  >  the  included  Z  of  the 
second,  the  third  side  of  the  first  is  >  the  third  side  of  the  second.] 

(§91) 

But  each  of  these  conclusions  is  contrary  to  the  hypothe- 
sis that  BC  is  >  EF. 

Then,  ii  ZA  can  be  neither  equal  to  Z  D,  nor  <  ZD, 

ZA>ZD. 


42  PLANE  GEOMETRY.  — BOOK  I. 


Prop.  XXX.     Tiikorem. 

93.  Tn  an  isosceles  triangle,  the  angles  opposite  the  equal 
sides  are  equal, 

c 


Given  AC  and  BC  the  equal  sides  of  isosceles  A  ABC* 
To  Prove  ZA^ZB, 

Proof.     Draw  line  CD  ±  AB, 
In  It.  A  AOD  iind  BCD, 

CD=CD, 
And  by  hyp.,  AC=BC 

,-.  AACD  =  ABCD. 
[Two  rt.  /S^  9,re  etjiial  when  the  hypotenuse  and  a  leg  of  one  are 
equal  reapeoUvely  to  the  hypotenuse  and  a  leg  of  the  other,]       (§  IH)) 

.'.  Z  .4  =  Z  5. 

[In  equal  figures,  the  homologous  pai'ta  are  equal.]  (§  GO) 

94.  Cor.  I.    From  equal  A  ACD  and  BCD,  we  have 

AD^^BDy  and  /LACD^ZBCDy  hence, 

1.  The  perpendicular  from  the  vertex  to  the  base  of  an 
iaoaoelea  triangle  bisects  the  base. 

2.  The  perpendicular  from  the  vertex  to  the  base  of  an 
isosceles  triangle  bisects  the  vertical  angle. 

95.  Cor.  II.    An  eq^tilateml  triangle  is  also  equiangular. 

Prop.  XXXI,    Theorem. 

96.  (Converse  of  Vvo\\  XXX.)    Iftivo  angles  of  a  triangle 

are  equal,  the  sides  opposite  are  equal. 


RECTILINEAR  FIGURES.  43 

(Fig.  of  Prop.  XXX.) 
Given,  in  A  ABC,  ZA  =  Z  B. 

To  Prove  AC  =  EC. 

(Prove  A  ACD  =  A  BCD  by  §  89.) 

97.  Cor.     An  equiangular  triangle  is  also  equilateral. 

EXERCISES. 

22.  The  angles  A  and  B  of  a  triangle  ABC  are  57°  and  98°  respec- 
tively ;  how  many  degrees  are  there  in  the  exterior  angle  at  C  ? 

23.  How  many  degrees  are  there  in  each  angle  of  an  equiangular 
triangle  ? 

Prop.  XXXII.    Theorem. 

98.  If  two  sides  of  a  triangle  are  unequal,  the  angles  oppo- 
site are  unequal,  and  the  greater  angle  lies  opposite  the  greater 
side. 

A 


Given,  in  A  ABC,         AC  >  AB. 
To  Prove  ZABC>ZC. 

Proof.     Take  AD  =  AB,  and  draw  line  BD. 
Then,  in  isosceles  A  ABD, 

ZABD  =  ZADB. 
[In  an  isosceles  A,  the  A  opposite  the  equal  sides  are  equal.]    (§  93) 
Now  since  Z  ADB  is  an  ext.  Z  of  A  BDG, 

Z  ADB  >ZC. 
[An  ext.  Z  of  a  A  is  >  either  of  the  opposite  int.  A.]  (§  86) 

Therefore,  its  equal,  Z  ABD,  is  >  ZC. 
Then,  since  Z  ABC  is  >  Z  ABD,  and  Z  ABD  >ZC, 

ZABOZa 


44  PLANE   GEOMETRY.— BOOK  I. 

Prop.  XXXIII.     Theorem. 

99.  (Converse  of  Prop.  XXXII.)  If  two  angles  of  a  tri- 
angle are  unequal,  the  sides  opposite  are  unequal,  and  the 
greater  side  lies  opposite  the  greater  angle. 

A 


Given,  in  A  ABC,  Z  ABC  >  Z  C. 
To  Prove  AC  >  AB. 

Proof.  Draw  line  BD,  making  Z  CBD  =  Z  C,  and  meet- 
ing AC  Sit  D. 

Then,  in  A  BCD,  BD  =  CD. 

[If  two  A  of  a,  A  are  equal,  the  sides  opposite  are  equal,]         (§  96) 

But,  AD  +  BD>AB. 

[A  str.  line  is  the  shortest  line  between  two  points.]  (Ax.  4) 

Substituting  for  BD  its  equal  CD,  we  have 

AD+CD>  AB,  or  AC>AB. 

Prop.  XXXIV.     Theorem. 

100.  If  straight  lines  he  drawn  from  a  point  within  a 
triangle  to  the  extremities  of  any  side,  the  angle  included  by 
them  is  greater  than  the  angle  included  by  the  other  two  sides. 


Given  D,  any  point  within  A  ABC,  and  lines  BD  and  CD. 


J 


RECTILINEAR  FIGURES. 

To  Prove  ZBDC>ZA. 

Proof.     Produce  BD  to  meet  AC  dJt  E. 
Then,  since  Z  BDC  is  an  ext.  Z  of  A  CDE, 

ZBDC>ZDEC. 

[An  ext.  Z  of  a  A  is  >  either  of  the  opposite  int.  zi.] 

In  like  manner,  since  Z  DEO  is  an  ext.  Z  of  A  ABE, 

ZDEOZA. 
Then,  since  Z  ^Z>(7  is  >  Z  DEC,  and  Z  2)^0  >  Z  ^ 

Zi?Z)0>ZA 


45 


(§85) 


Prop.  XXXV.    Theorem. 

101.  Any  point  in  the  bisector  of  an  angle  is  equally  distant 
from  the  sides  of  the  angle. 

/A 


Given  P,  any  point  in  bisector  BD  of  Z  ABC,  and  lines 
PJf  and  FN±  to  AB  and  AC,  respectively. 

To  Prove  PM  =  PiV: 

Proof.     In  rt.  A  BPM  and  BPN, 
BP=BP. 

And  by  hyp.,  Z  PBM=  Z  PBN. 

.-.  ABPM=ABPN. 

[Two  rt.  A  are  equal  when  the  hypotenuse  and  an  adj.  Z  of  one  are 
equal  respectively  to  the  hypotenuse  and  an  adj.  Z  of  the  other.] 

(§70) 
.-.  PM=PN. 


[In  equal  figures,  the  homologous  parts  are  equal] 


(§  ^>«) 


46 


PLANE   GEOMETRY. —BOOK  I. 


Prop.  XXXVI.     Theorem. 


102.  (Converse  of  Prop.  XXXV.)  Every  point  which  is 
within  an  angle,  and  equally  distant  from  its  sides,  lies  in  the 
bisector  of  the  angle. 

/A 


Given  point  P  within  Z  ABC,  equally  distant  from  sides 
AB  and  BC,  and  line  BP. 


To  Prove 


ZPBM=ZPBN. 


(Prove  ABPM  =  ABPNy  by  §  90;  the  theorem  then 
follows  by  §  QQ.) 

EXERCISES. 

24.  The  angle  at  the  vertex  of  an  isosceles  triangle  ABC  is  equal 
to  five-thirds  the  sum  of  the  equal  angles  B  and  C.  How  many  degrees 
are  there  in  each  angle  ? 

25.  If  from  a  point  0  in  a  straight  line  AB  lines  OC  and  OD  be 
drawn  on  opposite  sides  of  AB,  making  Z.AOC  =  Z.BOD,  prove 
that  OC  and  OD  lie  in  the  same  straight  line, 

(Fig.  of  Prop.  IV.  We  have  ZAOD  +  ZB0D  =  2  rt.  /i,  and  by 
hyp.,  ZBOD=  ZAOC.) 

26.  If  the  bisectors  of  two  adjacent  angles  make 
au  angle  of  45°  with  each  other,  the  angles  are  com- 
plementary. 

(Given  OD  and  0-E^  the  bisectors  of  AAOB  and 
BOC,  respectively,  and  ZDOE  =  i^° ;  to  prove 
A  AOB  and  BOC  complementary.) 

27.  Prove  Prop.  XXX.  by  drawing  CD  to  bisect  Z  ACB.  (§  63.) 

28.  Prove  Prop.  XXX.  by  drawing  CD  to  the  middle  point  of  AB 

29.  Prove  Prop.  XXXI.  by  drawing  CD  to  bisect  ZACB.  (§  68.) 


EECTILINEAR  FIGURES.  47 

QUADRILATERALS. 
DEFINITIONS. 

103.  A  quadrilateral  is  a  portion  of  a  plane  bounded  by- 
four  straight  lines ;  as  ABCD.  ^ 

The  bounding  lines  are  called  the  sides  /\. 

of  the  quadrilateral,  and  their  points  of         /  n. 

intersection  the  vertices.  /  N. 

The  angles  of  the  quadrilateral  are  the  ^\™ ""/C 

angles   included  between  the  adjacent  ^\^^  /^ 

sides.  D 

A  diagonal  is  a  straight  line  joining  two  opposite  vertices ; 
as  AO. 

104.  A  Trapezium  is  a  quadrilateral  no  two  of  whose  sides 
are  parallel. 

A  Trapezoid  is  a  quadrilateral  two,  and  only  two,  of 
whose  sides  are  parallel. 

A  Parallelogram  (O)  is  a  quadrilateral  whose  opposite 
sides  are  parallel. 


TVapezium.  Trapezoid.  Parallelogram. 

The  bases  of  a  trapezoid  are  its  parallel  sides;  the  alti- 
tude is  the  perpendicular  distance  between  them. 

If  either  pair  of  parallel  sides  of  a  parallelogram  be 
taken  and  called  the  hases,  the  altitude  corresponding  to 
these  bases  is  the  perpendicular  distance  between  them. 

105.  A  Rhomboid  is  a  parallelogram  whose  angles  are 
not  right  angles,  and  whose  adjacent  sides  are  unequal. 

A  Rhombus  is  a  parallelogram  whose  angles  are  not  right 
angles,  and  whose  adjacent  sides  are  equal. 

A  Rectangle  is  a  parallelogram  whose  angles  are  right 
angles. 


48  PLANE   GEOMETRY. —BOOK  I. 

A  Square  is  a  rectangle  whose  sides  are  equal. 


Rhoniboid. 


Rhombus. 


Rectangle. 


Square. 


Prop.  XXXVII.     Theorem. 
106.   In  any  parallelogram, 
I.    TJie  opposite  sides  are  equal. 
II.    Tlie  opposite  angles  are  equal. 

B, 


Given  0^5(7Z>. 

I.  To  Prove         AB=  CD  and  BC  =  AD. 
Proof.     Draw  diagonal  AC. 

In  A  ABC  and  ACD,  AC=AC. 

Again,  since  lis  BC  and  AD  are  cut  by  AC, 

Z  BCA  =  Z  CAD. 

[K  two  lis.  are  cut  by  a  transversal,  the  alt.  int.  A  are  equal.]  (§  72) 

.  In  like  manner,  since  lis  AB  and  CD  are  cut  by  AC, 

ZBAC=ZACD. 

.-.  A  ABC  =  A  ACD. 

[Two  A  are  equal  when  a  side  and  two  adj.  A  of  one  are  equal 
respectively  to  a  side  and  two  adj.  A  of  the  other.]  (§  68) 

.-.  AB  =  CD  and  BC  =  AD. 

[In  equal  figures,  the  homologous  parts  are  equal.]  (§  66) 

II.  To  Prove  Z BAD  =  Z BCD  Sind  ZB  =  ZD. 
Proof.     We  have  AB  II  CD,  and  AD  11  CB;  and  AB  and 

CD,  and  also  AD  and  CB,  extend  in  opposite  directions 
from  A  and  C. 


RECTILINEAR  FIGURES.  49 

.-.  Z  BAD  =  Z  BCD. 

[Two  A  whose  sides  are  ||,  each  to  each,  are  equal  if  both  pairs  of 
II  sides  extend  in  opposite  directions  from  their  vertices.]  (§  81) 

In  like  manner,  ZB  =  Z  D. 

107.  Cor.  I.     Parallel  lines  included  between  parallel  lines 
are  equal. 

108.  Cor.  n.    A  diagonal  of  a  parallelogram  divides  it 
into  two  equal  triangles. 

Prop.  XXXVIII.     Thj:orem. 

109.  (Converse  of  Prop.  XXXVII,  I.)    If  the  opposite  sides 
of  a  quadrilateral  are  equal,  the  figure  is  a  parallelogram. 


Given,  in  quadrilateral  ABCD, 

AB=CD  SindBC  =  AD. 
To  Prove  ABCD  a  O. 
Proof.     Draw  diagonal  AC. 
In  A  ABC  and  ACD,  AC  =  AC. 
And  by  hyp.,   AB  =  CD  and  BC  =  AD. 

.:  A  ABC  =  A  ACD. 
[Two  A  are  equal  when  the  three  sides  of  one  are  equal  respec- 
tively to  the  three  sides  of  the  other.]  (§  69) 

.-.  ZBCA  =  ZCAD3indZBAC=ZACD. 

[In  equal  figures,  the  homologous  parts  are  equal.]  (§  Q6) 

Since  Z  BCA  =  Z  CAD,  BC  II  AD. 

[If  two  str.  lines  are  cut  by  a  transversal,  and  the  alt.  int.  A  are 
equal,  the  two  lines  are  jj.]  (§  73) 

In  like  manner,  since  Z  BAC  =  Z  ACD,  AB  II  CD. 
Then  by  del,  ABCD  is  a  O. 


50  PLANE  GEOMETRY.— BOOK  I. 

Ex.  30.   If  one  angle  of  a  parallelogram  is  119°,  how  many  degrees 
are  there  in  each  of  the  others  ? 

Prop.  XXXIX.     Theorem. 

110.  If  two  sides  of  a  quadrilateral  are  equal  and  parallel, 
the  Jigure  is  a  parallelogram. 


Given,  in  quadrilateral  ABCD,  BC  equal  and  II  to  AD. 

To  Prove  ABOD  a  O. 

(Prove  AABC=AACD,  by  §63;  then,  the  other  two 
sides  of  the  quadrilateral  are  equal,  and  the  theorem  follows 
by  §  109.) 

Prop.  XL.     Theorem. 
HI.    TJie  diagonals  of  a  parallelogram  bisect  each  other. 

Br^. ^ ^-^^C 


Given  diagonals  AC  and  BD  of  O  ABCD  intersecting 
at  ^. 

To  Prove  AE  =  EC  and  BE  =  ED. 

(Prove  A  AED  =  A  BEC,  by  §  68.) 

Note.    The  point  E  is  called  the  centre  of  the  parallelogram. 

Prop.  XLI.     Theorem. 

112.   (Converse  of  Prop.  XL.)      If  the  diagonals  of  a 
quadrilateral  bisect  each  other,  the  figure  is  a  parallelogram. 


RECTILINEAR  FIGURES.  51 

(Fig.  of  Prop.  XL.) 

Given  AG  and  BD,  the  diagonals  of  quadrilateral  ABCD, 
bisecting  each  other  at  E. 

To  Prove  ABCD  a  O. 

(Frove  A  AED  =  A  BEC,  by  §  63;  then  AD  =  BC]  in 
like  manner,  AB=  CD,  and  the  theorem  follows  by  §  109.) 

Prop.  XLII.     Theorem. 

113.  Two  parallelograms  are  equal  when  two  adjacent  sides 
and  the  included  angle  of  one  are  equal  respectively  to  two 
adjacent  sides  and  the  included  angle  of  the  other. 


Given,  in  [U  ABCD  and  EFGH, 

AB  =  EF,  AD  =  EH,  and  ZA  =  ZE. 

To  Prove  CJABCD  =  CJ  EFGH. 

Proof.  Superpose  CJ  ABCD  upon  CJ  EFGH  in  such  a 
way  that  Z  A  shall  coincide  with  its  equal  Z  E ;  side  AB 
falling  on  side  EF,  and  side  AD  on  side  EH. 

Then  since  AB  =  EF  and  AD  =  EH,  point  B  will  fall  on 
point  F,  and  point  D  on  point  H 

Now  since  BC  II  AD  and  FG  II  EH,  side  BC  will  fall  on 
side  FG,  and  point  G  will  fall  somewhere  on  FG. 

[But  one  str.  line  can  be  drawn  through  a  given  point  jj  to  a  given 
str.  line.]  (§  53) 

In  like  manner,  side  DC  will  fall  on  side  HG,  and  point 
C  will  fall  somewhere  on  HG. 

Then  point  C,  falling  at  the  same  time  on  FG  and  HG, 
must  fall  at  their  intersection  G. 

Hence,  the  CEJ  coincide  throughout,  and  are  equal. 


52 


PLANE   GEOMETRY.— BOOK  I. 


114.  Cor.  Tivo  rectangles  are  equal  if  the  base  and  alti- 
tude of  one  are  equal  respectively  to  the  base  and  altitude  of 
the  other. 

Prop.  XLIII.    Theorem. 

115.  The  diagonals  of  a  rectangle  are  equal. 
B^ ^C 


Given  AC  and  BD  the  diagonals  of  rect.  ABCD, 
To  Prove  AC  =  BD. 

(Prove  rt.  A  ABD  =  rt.  A  ACD,  by  §  63.) 

116.  Cor.     The  diagonals  of  a  square  are  equal. 

Prop.  XLIV.     Theorem. 

117.  The  diagonals  of  a  rhombus  bisect  each  other  at  right 
angles.  ^ 


{AC  and  BD  bisect  each  other  at  rt.  z§  by  §  43.) 


EXERCISES. 

31.  The  bisector  of  the  vertical  angle  of  an  isosceles  triangle 
bisects  the  base  at  right  angles. 

(Fig.  of  Prop.  XXX.  In  equal  A  ACD  and  BCD,  we  have 
ZADC  =  ZBDC;  then  CD±AB  by  §  24.) 

32.  The  line  joining  the  vertex  of  an  isosceles  triangle  to  the 
middle  point  of  the  base,  is  perpendicular  to  the  base,  and  bisects  the 
vertical  angle. 

(Fig.  of  Prop.  XXX.     Prove  CD  ±  AB  as  in  Ex.  31.) 

33.  If  one  angle  of  a  parallelogram  is  a  right  angle,  the  figure  is  a 
rectangle. 


RECTILINEAR  FIGURES. 


63 


POLYGONS. 

DEFINITIONS. 

118.  A  polygon  is  a  portion  of  a  plane  bounded  by  three 
or  more  straight  lines ;  as  ABCDE. 

The  bounding  lines  are  called  the 
sides  of  the  polygon,  and  their  sum  is 
called  the  perimeter. 

The   angles  of  the   polygon  are  the 
angles  EAB,  ABC,  etc.,  included   be- 
tween the  adjacent  sides ;  and  their  vertices  are  called  the 
vertices  of  the  polygon. 

A  diagonal  of  a  polygon  is  a  straight  line  joining  any  two 
vertices  which  are  not  consecutive ;  as  AC. 

119.  Polygons  are  classified  with  reference  to  the  number 
of  their  sides,  as  follows : 


No.  or 
Sides. 

Designation. 

No.   OF 
Sides. 

Designation. 

3 
4 

5 
6 

7 

Triangle. 

Quadrilateral. 

Pentagon. 

Hexagon. 

Heptagon. 

8 

9 

10 

11 

12 

Octagon. 

Enneagon. 

Decagon. 

Hendecagon. 

Dodecagon. 

120.  An  equilateral  polygon  is  a  polygon  all  of  whose 
sides  are  equal. 

An  equiangular  polygon  is  a  polygon  all  of  whose  angles 
are  equal. 

121.  A  polygon  is  called  convex  when 
no  side,  if  produced,  will  enter  the  surface 
enclosed  by  the  perimeter ;  as  ABCDE. 

It  is  evident  that,  in  such  a  case,  each 
angle  of  the  polygon  is  less  than  two 
right  angles. 


54 


PLANE   GEOMETRY.— BOOK   I. 


All  polygons  considered  hereafter  will  be  understood  to 
be  convex,  unless  the  contrary  is  stated. 

A  polygon  is  called  concave  when  at  least  two  of  its  sides, 
if  produced,  will  enter  the  surface  enclosed        G.  ^ 

by  the  perimeter ;  as  FGHIK.  /    \n/^ 

It  is  evident  that,  in  such  a  case,  at  least        /  \ 

one  angle  of  the  polygon  is  greater  than       I \ 

two  right  angles.  -KT 

Thus,  in  polygon  FGHIK,  the  interior  angle  G HI  is 
greater  than  two  right  angles. 

SucK  an  angle  is  called  re-entrant. 

122.  Two  polygons  are  said  to  be  mutually  equilateral 

when  the  sides  of   one  are  p/ 

C  ' 

equal     respectively    to    the  ■ — V  B, 

sides    of    the    other,    when 

taken  in  the  same  order.  ^ 

Thus,     polygons     ABCD        ^  ^     A 

and  A'B'C'D'  are  mutually  equilateral  if 

AB  =  A'B',  BC  =  B'C',  GD=OD\  and  DA  =  D'A\ 

Two  polygons  are  said  to  be  mutually  equiangular  when 
the    angles    of   one    are   equal 
respectively   to   the   angles   of 
the   other  when   taken  in  the  F 
same  order. 

Thus,  polygons  EFGH  and       ^ 
E'F'G'H'  are  mutually  equiangular  if 

ZE=ZE',  ZF=ZF',  AG  =  ZG',  and  ZH=ZH'. 

123.  In  polygons  which  are  mutually  equilateral  or 
mutually  equiangular,  sides  or  angles  which  are  similarly 
placed  are  called  homologous. 

In  mutually  equiangular  polygons,  the  sides  included 
between  equal  angles  are  homologous. 


124.   If  two  triangles  are  mutually  equilateral,  they  are 
also  mutually- equiangular  (§  69). 


I 


RECTILINEAR  FIGURES.  55 

But  with  this  exception,  two  polygons  may  be  mutually 
equilateral  without  being  mutually  equiangular,  or  mutually 
equiangular  without  being  mutually  equilateral. 

If  two  polygons  are  both  mxUually  equilateral  and  mutually 
equiangular,  they  are  equal. 

For  they  can  evidently  be  applied  one  to  the  other  so  as 
to  coincide  throughout. 

125.  Two  polygons  are  equal  when  they  are  composed  of  the 
same  number  of  triangles,  equal  ea^h  to  each,  and  similarly 
placed. 

For  they  can  evidently  be  applied  one  to  the  other  so  as 
to  coincide  throughout. 

Prop.  XLV.     Theorem. 

126.  The  sum  of  the  angles  of  any  polygon  is  equal  to  two 
right  angles  taken  as  many  times,  less  two,  as  the  polygon  has 
sides. 


Given  a  polygon  of  n  sides. 

To  Prove  the  sum  of  its  A  equal  to  n  —  2  times  two  rt.  A. 

Proof.  The  polygon  may  be  divided  into  n  —  2  A  by 
drawing  diagonals  from  one  of  its  vertices. 

The  sum  of  the  A  of  the  polygon  is  equal  to  the  sum  of 
the  A  of  the  A. 

But  the  sum  of  the  A  of  each  A  is  two  rt.  A. 

[The  sum  of  the  A  of  any  A  is  equal  to  two  rt.  A.'\  (§  84) 

Hence,  the  sum  of  the  A  of  the  polygon  is  n  —  2  times 
two  rt.  A. 


56  PLANE  GEOMETRY.— BOOK  I. 

127.  Cor.  I.  The  sum  of  the  angles  of  any  polygon  is  equal 
to  twice  as  many  right  angles  as  the  polygon  has  sides,  less 
four  right  angles. 

For  if  B  represents  a  rt.  Z,  and  n  the  number  of  sides  of 
a  polygon,  the  sum  of  its  A  is  (ri  —  2)x2E,  or  2  nR  —  ^B. 

128.  Cor.  II.  The  sum  of  the  angles  of  a  quadrilateral  is 
equal  to  four  right  angles;  of  a  pentagon,  six  right  angles;  of 
a  hexagon,  eight  right  angles;  etc. 

Prop.  XLVI.     Theorem. 

129.  If  the  sides  of  any  polygon  be  produced  so  as  to  make 
an  exterior  angle  at  each  vertex,  the  sum  of  these  exterior 
angles  is  equal  to  four  right  angles. 


Given  a  polygon  of  n  sides  with  its  sides  produced  so  as 
to  make  an  ext.  Z  at  each  vertex. 

To  Prove  the  sum  of  these  ext.  A  equal  to  4  rt.  A. 

Proof.     The  sum  of  the  ext.  and  int.  A  at  any  one  vertex 
is  two  rt.  A. 

[If  two  adj.  A  have  their  ext.  sides  in  the  same  str.  line,  their  sum 
is  equal  to  two  rt.  A.']  (§  32) 

Hence,  the  sum  of  all  the  ext.  and  int.  A\^2n  rt.  A. 
But  the  sum  of  the  int.  A  alone  is  2  n  rt.  zi  —  4  rt.  A. 

[The  sum  of  the  A  of  any  polygon  is  equal  to  twice  as  many  rt.  A 
as  the  polygon  has  sides,  less  4  rt.  zi.]  (§  127) 

Whence,  the  sum  of  the  ext.  z^  is  4  rt.  A 


RECTILINEAR  FIGURES.  57 


EXERCISES. 

34.  How  many  degrees  are  there  in  each  angle  of  an  equiangular 
hexagon  ?  of  an  equiangular  octagon  ?  of  an  equiangular  decagon  ? 
of  an  equiangular  dodecagon  ? 

35.  How  many  degrees  are  there  in  the  exterior  angle  at  each 
vertex  of  an  equiangular  pentagon  ? 

36.  If  two  angles  of  a  quadrilateral  are  supplementary,  the  other 
two  angles  are  supplementary. 

37.  If,  in  a  triangle  ABC,  ZA=ZB,  a  line  par- 
allel to  AB  makes  equal  angles  with  sides  AC  and 
BC. 

(To  prove  Z  CDE  =  Z  CED.) 

38.  If  the  equal  sides  of  an  isosceles  triangle  be 
produced,  the  exterior  angles  made  with  the  base 
are  equal.     (§31,2.)  ^/ \b 

d/  \e 

39.  If  the  perpendicular  from  the  vertex  to  the  base  of  a  triangle 
bisects  the  base,  the  triangle  is  isosceles. 

(Fig.  of  Prop.  XXX.    A  ACD  and  BCD  are  equal  by  §  63.) 

\C 

40.  The  bisectors  of  the  equal  angles  of  an  isosceles 
triangle  form,  with  the  base,  another  isosceles  triangle. 

41.  If  from  any  point  in  the  base  of  an  isosceles  tri- 
angle perpendiculars  to  the  equal  sides  be  drawn,  they 
make  equal  angles  with  the  base.  Ey 

(ZADE  =  ZBDF,  by  §  31,  1.)  A- 

D 

42.  If  the  angles  adjacent  to  one  base  of  a        b, -yC 

trapezoid  are  equal,  those  adjacent  to  the  other         /  \ 

base  are  also  equal.  / 

(Given  ZA  =  ZD;  to  prove  ZB  =  ZC.)  ^  

43.  Either  exterior  angle  at  the  base  of  an  isos- 
celes triangle  is  equal  to  the  sum  of  a  right  angle 
and  one-half  the  vertical  angle. 

U  DAE  is  an  ext.  Z  of  A  ACD.) 

E 


58 


PLANE   GEOMETRY. —BOOK  I. 


44.  The  straight  lines  bisecting  the  equal  angles 
of  an  isosceles  triangle,  and  terminating  in  the  oppo- 
site sides,  are  equal. 

45.  Two  isosceles  triangles  are  equal  when  the  base  and  vertical 
angle  of  one  are  equal  respectively  to  the  base  and  vertical  angle  of 
the  other. 

(Each  of  the  remaining  A  of  one  A  is  equal  to  each  of  the  remain- 
ing A  of  the  other.) 

46.  If  two  parallels  are  cut  by  a  transversal,      ^^  e. 
the  bisectors  of  the  four  interior  angles  form  a 
rectangle.                                                                          H^ 

(EHW  FG,  by  §  73  ;  in  like  manner,  EF  \\  GH;     c 
then  use  Exs.  12  and  33.) 

47.  Prove  Prop.  XXVI.  by  drawing  through 
B  a  line  parallel  to  AC. 

(Sum  oiA&tB  =  2  rt.  A.) 


MISCELLANEOUS   THEOREMS. 

Prop.  XLVII.     Theorem. 

130.  The  line  joining  the  middle  points  of  two  sides  of  a 
triangle  is  parallel  to  the  third  side,  and  equal  to  one-half 
of  it. 


Given  line  DE  joining  middle  points  of  sides  AB  and 
AC,  respectively,  of  A  ABC. 

To  Prove  DE  11  BC,  and  DE  =  ^BC. 

Proof.     Draw  line  BF  II  AC,  meeting  ED  produced  at  F. 


RECTILINEAR  FIGURES.  59 

In  A  ADE  and  BDF, 

Z  ADE  =  Z  BDF. 

[If  two  str.  lines  intersect,  the  vertical  A  are  equal.]  (§  40) 

Also,  since  lis  AC  and  BF  are  cut  by  AB, 

ZA  =  Z  DBF. 

[If  two  lis  are  cut  by  a  transversal,  the  alt.  int.  A  are  equal.]  (§  72) 

And  by  hyp.,  AD  =  BD. 

.-.  A  ADE  =  A  BDF. 

[Two  A  are  equal  when  a  side  and  two  adj.  A  of  one  are  equal  re- 
spectively to  a  side  and  two  adj.  A  of  the  other.]  (§  68) 

.-.  DE  =  DF  and  AE  =  BF. 

[In  equal  figures,  the  homologous  parts  are  equal.]  (§  66) 

Then  since,  by  hyp.,  AE  —  EC,  BF  is  equal  and  II  to  CE. 

Whence,  BCEF  is  a  O. 

[If  two  sides  of  a  quadrilateral  are  equal  and  ||,  the  figure  is  a  O.] 

(§  110) 
.-.  DEWBC. 

Again,  since  DE  =  DF, 

DE  =  \FE  =  \BC. 

[In  any  O,  the  opposite  sides  are  equal.]  (§  106) 

131.   Cor.     The  line  which  bisects  one  side  of  a  triangle, 
and  is  parallel  to  another-  side,  bisects  also 
the  third  side. 

Given,  in  A  ABC,  D  the  middle  point 
of  side  AB,  and  line  DE  II  BC. 

To  Prove  that  DE  bisects  AC. 

Proof.     A  line  joining  D  to  the  middle 
point  of  AC  will  be  11  BC. 

[The  line  joining  the  middle  points  of  two  sides  of  a  A  is  ||  to  the 
third  side.]  (§  130) 

Then  this  line  will  coincide  with  DE. 

[But  one  str.  line  can  be  drawn  through  a  given  point  ||  to  a  given 
str.  line.]  (§  53) 

Therefore,  DE  bisects  AC. 


60  PLANE   GEOMETRY. —BOOK  I. 

Prop.  XLVIII.     Theorem. 

132.  TJie  line  joining  the  middle  points  of  the  non-parallel 
sides  of  a  trapezoid  is  parallel  to  the  bases,  and  equal  to  one- 
half  their  sum. 


Given  line  EF  joining  middle  points  of  non-ll  sides  AB 
and  CD,  respectively,  of  trapezoid  ABCD. 

To  Prove  EF  II  to  AD  and  EC,  and  EF  =  \  {AD  +  BC). 

Proof.  If  EF  is  not  II  to  AD  and  BC,  draw  line  EK 
II  to  AD  and  BC,  meeting  CD  at  K-,  and  draw  line  BD  in- 
tersecting EF  at  G,  and  EK  at  H. 

In  A  ABD,  EH  is  II  AD  and  bisects  AB ;  then  it  bisects 
BD. 

[The  line  which  bisects  one  side  of  a  A,  and  is  |I  to  another  side, 
bisects  also  the  third  side.]  (§  131) 

In  like  manner,  in  ABCD,  HK  is  II  BC  and  bisects  BD; 
then  it  bisects  CD. 

But  this  is  impossible  unless  EK  coincides  with  EF. 

[But  one  str.  line  can  be  drawn  between  two  points.]  (Ax.  3) 

Hence,  EF  is  II  to  AD  and  BC. 

Again,  since  EG  coincides  with  EH,  and  EH  bisects  AB 

^nd  BD,  EG  =  \AD.  (1) 

[The  line  joining  the  middle  points  of  two  sides  of  a  A  is  equal  to 
one-half  the  third  side.]  (§  130) 

In  like  manner,  since  GF  bisects  BD  and  CD, 

GF=\BC.  (2) 

Adding  (1)  and  (2), 

EG+GF=^AD  +  ^BG. 
Or,  EF=\{AD+BC). 


RECTILINEAR   FIGURES.  61 

133.  Cor.  The  line  which  is  parallel  to  the  bases  of  a  traj)- 
ezoid,  and  bisects  one  of  the  non-parallel  sides,  bisects  the  other 
also. 

Prop.  XLIX.     Theorem. 

134.  TJie  bisectors  of  the  angles  of  a  triangle  intersect  at 
a  common  point. 

C 


Given  lines  AD,  BE,  and  CF  bisecting  A  A,  B,  and  0, 
respectively,  of  A  ABC. 

To  Prove  that  AD,  BE,  and  CF  intersect  at  a  common 
point. 

Proof.     Let  AD  and  BE  intersect  at  0. 
Since  0  is  in  bisector  AD,  it  is  equally  distant  from  sides 
AB  and  AC. 

[Any  point  in  the  bisector  of  an  Z  is  equally  distant  from  the  sides 
of  the  Z.]  (§  101) 

In  like  manner,  since  0  is  in  bisector  BE,  it  is  equally 
distant  from  sides  AB  and  BC. 

Then  0  is  equally  distant  from  sides  AC  and  BC,  and 
therefore  lies  in  bisector  CF. 

[Every  point  which  is  within  an  Z,  and  equally  distant  from  its 
sides,  lies  in  the  bisector  of  the  Z.]  (§  102) 

Hence,  AD,  BE,  and  CF  intersect  at  the  common  point  0. 

135.  Cor.  TJie  point  of  intersection  of  the  bisectors  of 
the  angles  of  a  triangle  is  equally  distant  from  the  sides  of 
the  triangle. 


62 


PLANE   GEOMETRY.  — BOOK  I. 


Prop.  L.     Theorem. 

136.    The  perpendiculars  erected  at  the  middle  points  of  the 
sides  of  a  triangle  intersect  at  a  common  point. 


Given  DG,  EH,  and  FK  the  Js  erected  at  middle  points  D, 
E,  and  F,  of  sides  BC,  GA,  and  AB,  respectively,  of  A  ABC. 

To  Prove  that  DG,  EH,  and  FK  intersect  at  a  common 
point. 

(Let  DG  and  EH  intersect  at  0 ;  by  §  41,  0  is  equally 
distant  from  B  and  C;  it  is  also  equally  distant  from  A 
and  (7;  the  theorem  follows  by  §  42.) 

137.  Cor.  The  point  of  intersection  of  the  perpendiculars 
erected  at  the  middle  poirits  of  the  sides  of  a  triangle,  is  equally 
distant  from  the  vertices  of  the  triangle. 

EXERCISES. 

48.  If  the  diagonals  of  a  parallelogram  are  equal,  the  figure  is  a 
rectangle. 

(Fig.  of  Prop.  XLIII.  A  ABD  and  ACD  are  equal,  and  therefore 
ZBAJ)=/.ADC ;  also,  these  A  are  supplementary.) 

49.  If  two  adjacent  sides  of  a  quadrilateral  are 
equal,  and  the  diagonal  bisects  their  included  angle, 
the  other  two  sides  are  equal, 

(Given  AB  =  AD,  and  AC  bisecting  Z  BAD;  to 
prove  BC=  CD.) 

I 

50.  The  bisectors  of  the  interior  angles  of  a 

parallelogram  form  a  rectangle, 

(By  Ex.  46,  each  Z  of  EFGH is  a  rt.  Z.)         /^        h 


RECTILINEAR  FIGURES. 


63 


Prop.  LI.     Theorem. 

138.    The  perpendiculars  from  the  vertices  of  a  triangle  to 
the  opposite  sides  intersect  at  a  common  point. 


K^ 


Given  AD,  BE,  and  Ci^the  Js  from  the  vertices  of  A  ABC 
to  the  opposite  sides. 

To  Prove  that  AD,  BE,  and  CF  intersect  at  a  common 
point. 

Proof.     Through  A,  B,  and  C,  draw  lines  HK,  KG,  and 
GH  II  to  BC,  CA,  and  AB,  respectively,  forming  A  GHK 
Then  AD,  being  i.  BC,  is  also  ±  HK. 

[A  str.  line  ±  to  one  of  two  ||s  is  ±  to  the  other.]  (§  56) 

Now  since,  by  cons.,  ABCH  a,nd  ACBK&ve  UJ, 

AH=BC  and  AK=Ba 
[In  any  O,  the  opposite  sides  are  equal.]  (§  106) 

.-.  AH=AK 

[Things  which  are  equal  to  the  same  thing,  are  equal  to  each  other.] 

(Ax.  1) 

Then  AD  is  i.  fiT/f  at  the  middle  point  of  HK 

In  like  manner,  BE  and  CF  are  _L  to  KG  and  GH,  respec- 
tively, at  their  middle  points. 

Then,  AD,  BE,  and  CF  being  i.  to  the  sides  of  A  GHK 
at  their  middle  points,  intersect  at  a  common  point. 

[The  ±  erected  at  the  middle  points  of  the  sides  of  a  A  intersect 
at  a  common  point.]  (§1^6) 


64  PLANE   GEOMETRY.  — BOOK  I. 

139.  Def.  A  median  of  a  triangle  is  a  line  drawn  from 
any  vertex  to  the  middle  point  of  the  opposite  side. 

Prop.  LII.     Theorem. 

140.  The  medians  of  a  triangle  intersect  at  a  common  pointy 
which  lies  two-thirds  the  way  from  each  vertex  to  the  middle 
point  of  the  opposite  side. 

,     C 


A  F  B 

Given  AD,  BE,  and  CF  the  medians  of  A  ABC. 

To  Prove  that  AD,  BE,  and  CF  intersect  at  a  common 
point,  which  lies  two-thirds  the  way  from  each  vertex  to  the 
middle  point  of  the  opposite  side. 

Proof.     Let  AD  and  BE  intersect  at  0. 

Let  G  and  H  be  the  middle  points  of  OA  and  OB,  respec- 
tively, and  draw  lines  ED,  OH,  EG,  and  DH. 

Since  ED  bisects  AC  and  BC, 

ED\\AB2.n&  =\AB. 

[The  line  joining  the  middle  points  of  two  sides  of  a  A  is  ||  to  the 
third  side,  and  equal  to  one-half  of  it.]  (§  130) 

In  like  manner,  since  GH  bisects  OA  and  OB, 

GH  11  AB  and  =  i  AB. 
Then  ED  and  GH  are  equal  and  II. 

[Things  which  are  equal  to  the  same  thing,  are  equal  to  each  other.] 

(Ax.  1) 
[Two  str.  lines  11  to  the  same  str.  line  are  ||  to  each  other.]  (§  55) 
Therefore,  EDHG  is  a  O. 

[If  two  sides  of  a  quadrilateral  are  equal  and  ||,  the  figure  is  a  O.] 

(§  110) 


RECTILINEAR  FIGURES.  65 

Then  GD  and  EH  bisect  each  other  at  0. 

[The  diagonals  of  a  O  bisect  each  other.]  (§  HI) 

But  by  hyp.,  O  is  the  middle  point  of  OA,  and  H  of  OB. 
.-.  AG=OG=  OD,  and  BH=  0H=  OE. 

That  is,  AD  and  BE  intersect  at  a  point  0  which  lies 
two-thirds  the  way  from  A  to  D,  and  from  B  to  ^. 

In  like  manner,  AD  and  CF  intersect  at  a  point  which 
lies  two-thirds  the  way  from  A  to  D,  and  from  C  to  F. 

Hence,  AD,  BE,  and  CF  intersect  at  the  common  point 
0,  which  lies  two-thirds  the  way  from  each  vertex  to  the 
middle  point  of  the  opposite  side. 

LOCI. 

141.  Def.  If  a  series  of  points,  all  of  which  satisfy  a 
certain  condition,  lie  in  a  certain  line,  and  every  point  in 
this  line  satisfies  the  given  condition,  the  line  is  said  to  be 
the  locus  of  the  points. 

For  example,  every  point  which  satisfies  the  condition  of 
being  equally  distant  from  the  extremities  of  a  straight  line, 
lies  in  the  perpendicular  erected  at  the  middle  point  of  the 
line  (§  42). 

Also,  every  point  in  the  perpendicular  erected  at  the 
middle  point  of  a  line  satisfies  the  condition  of  being 
equally  distant  from  the  extremities  of  the  line  (§  41). 

Hence,  the  perpeyidicular  erected  at  the  middle  point  of  a 
straight  line  is  the  Locus  of  points  which  are  equally  distant 
from  the  extremities  of  the  line. 

Again,  every  point  which  satisfies  the  condition  of  being 
within  an  angle,  and  equally  distant  from  its  sides,  lies  in 
the  bisector  of  the  angle  (§  102). 

Also,  every  point  in  the  bisector  of  an  angle  satisfies  the 
condition  of  being  equally  distant  from  its  sides  (§  101). 

Hence,  the  bisector  of  an  angle  is  the  locus  of  points  which 
are  within  the  angle,  and  equally  distant  from  its  sides. 


66  PLANE   GEOMETRY.— BOOK  I. 

EXERCISES. 

51.  Two  straight  lines  are  parallel  if  any  two  points  of  either 
are  equally  distant  from  the  other. 

(Prove  by  Beductio  ad  Absurdum.) 

52.  What  is  the  locus  of  points  at  a  given  distance  from  a  given 
straight  line  ?    (Ex.  51.) 

53.  What  is  the  locus  of  points  equally  distant  from  a  pair  of 
intersecting  straight  lines  ? 

54.  What  is  the  locus  of  points  equally  distant  from  a  pair  of 
parallel  straight  lines  ? 

D  n 

55.  The  bisectors  of  the  interior  angles  of  a 
trapezoid  form  a  quadrilateral,  two  of  whose 
angles  are  right  angles.     (Ex.  46.) 


56.  If  the  angles  at  the  base  of  a  trapezoid  B^ n 

are  equal,  the  non-parallel  sides  are  also  equal.  /-X            \ 

(Given  ZA  =  ZD;  to  prove  AB  =  CD.   Draw  /      \            \ 

BEWCD.)  ^^ i ^D 

57.  If  the  non-parallel  sides  of  a  trapezoid  are  equal,  the  angles 
which  they  make  with  the  bases  are  equal. 

(Fig.  of  Ex.  56.    Given  AB=  CD;  to  prove  Z^  =  ZD,  and  also 
ZABC=ZC.    DrsiW  BE  \\  CD.) 

58.  The  perpendiculars  from  the  extremities  of 
the  base  of  an  isosceles  triangle  to  the  opposite  sides 
are  equal. 

59.  If  the  perpendiculars  from  the  extremities  of  the  base  of  a 
triangle  to  the  opposite  sides  are  equal,  the  triangle  is  isosceles. 

(Converse  of  Ex.  58.    Prove  AACD  =  ABCE.) 

60.  The  angle  between  the  bisectors  of  the  equal 
angles  of  an  isosceles  triangle  is  equal  to  the  exte- 
rior angle  at  the  base  of  the  triangle. 

(Z  ADB  =  180°  - (Z  BAD  +  Z  ABD).) 

61.  If  a  line  joining  two  parallels  be  bisected, 
any  line  drawn  through  the  point  of  bisection 
and  included  between  the  parallels  will  be  bisected 
at  the  point. 

(To  prove  that  GH  is  bisected  at  0.) 


RECTILINEAR  FIGURES. 


67 


62.  If  through  a  point  midway  between  two 
parallels  two  transversals  be  drawn,  they  inter- 
cept equal  portions  of  the  parallels. 

(Draw  0K±  AB,  and  produce  KO  to  meet  CD 
at  L.     Then  A  OGK  =  A  OIIL.) 

63.  If  perpendiculars  BE  and  DF  be  drawn 
from  vertices  B  and  D  of  parallelogram  ABCD 
to  the  diagonal  AC,  prove  BE  =  DF.     (§  70.) 


64.  The  lines  joining  the  middle  points  of  the 
sides  of  a  triangle  divide  it  into  four  equal  trian- 
gles.    (§130.) 


65.  If  from  any  point  in  the  base  of  an  isosceles 
triangle  parallels  to  the  equal  sides  be  drawn,  the 
perimeter  of  the  parallelogram  formed  is  equal  to  the 
sum  of  the  equal  sides  of  the  triangle.     (§  96.) 


66.  The  bisector  of  the  exterior  angle  at  the  ver- 
tex of  an  isosceles  triangle  is  parallel  to  the  base. 
(§  85,  1.) 


67.   The  medians  drawn  from  the  extremities  of 
the  base  of  an  isosceles  triangle  are  equal. 


68.  If  from  the  vertex  of  one  of  the  equal  angles 
of  an  isosceles  triangle  a  perpendicular  be  drawn  to 
the  opposite  side,  it  makes  with  the  base  an  angle 
equal  to  one-half  the  vertical  angle  of  the  triangle. 

(To  prove  Z  BAD  =  ^ZC.) 


69.  If  the  exterior  angles  at  the  vertices  A 
and  B  of  triangle  ABC  are  bisected  by  lines 
vsrhich  meet  at  D,  prove 

Z  ADB  =  90°  -  1  C. 

(Z  ADB  =  180°  -  (Z  BAD  +  Z  ABD) .) 


68 


PLANE   GEOMETRY.— BOOK  I. 


70.  The  diagonals  of  a  rhombus  bisect  its  angles. 
(Fig.  of  Prop.  XLIV.) 

71.  If  from  any  point  in  the  bisector  of  an  angle 
a  parallel  to  one  of  the  sides  be  drawn,  the  bisector, 
the  parallel,  and  the  remaining  side  form  an  isosceles 
triangle. 

72.  If  the  bisectors  of  the  equal  angles  of  an  isos- 
celes triangle  meet  the  equal  sides  at  B  and  E,  prove 
DE  parallel  to  the  base  of  the  triangle. 

(Prove  A  CED  isosceles.) 

73.  If  at  any  point  D  in  one  of  the  equal  sides 
AB  of  isosceles  triangle  ABC,  DE  be  drawn  per- 
pendicular to  base  BC  meeting  CA  produced  at  E, 
prove  triangle  ADE  isosceles. 

74.  From  C,  one  of  the  extremities  of  the  base 
^C  of  isosceles  triangle  ABC,  a  line  is  drawn  meet- 
ing BA  produced  at  D,  making  AD  =  AB.  Prove 
CD  perpendicular  to  BC.     (§  84.) 

(A  ACD  is  isosceles.) 

75.  If  the  non-parallel  sides  of  a  trapezoid 
are  equal,  its  diagonals  are  also  equal.     (Ex.  57.) 


76.  If  ADC  is  a  re-entrant  angle  of  quadrilateral 
ABCD,  prove  that  angle  ADC,  exterior  to  the  fig- 
ure, is  equal  to  the  sum  of  interior  angles  A,  B, 
and  C.     (§  128.)  B 

77.  If  a  diagonal  of  a  quadrilateral  bisects  two 
of  its  angles,  it  is  perpendicular  to  the  other  diagonal,    a 

(Prove  AC±DB,  by  §  43.) 


78.  In  a  quadrilateral  ABCD,  angles  ABD 
and  CAD  are  equal  to  ACD  and  BDA,  respec- 
tively ;  prove  BC  parallel  to  AD. 

(Prove  AB=  CD;  then  prove  BE  =  CF.) 


79.   State  and  prove  the  converse  of  Prop.  XLIV.     (J  41,  I.) 


RECTILINEAR   FIGURES, 


69 


80.   State  and  prove  the  converse  of  Ex.66,  p.  67.     (§  96.) 


81.  The  bisectors  of  the  exterior  angles  at  two 
vertices  of  a  triangle,  and  the  bisector  of  the  inte- 
rior angle  at  the  third  vertex  meet  at  a  common 
point. 

(Prove  as  in  §  134.) 


82.  A  BCD  is  a  trapezoid  whose  parallel  sides 
AD  and  BG  are  perpendicular  to  CD.  If  E  is 
the  middle  point  of  AB,  prove  EC=  ED.  (§  41 ,  T.) 

(Draw^jPMZ).)  A 


83.  The  middle  point  of  the  hypotenuse  of  a 
right  triangle  is  equally  distant  from  the  vertices 
of  the  triangle. 

(To  prove  AD=BD=  CD.  Draw  DE\\BC.) 


84.   The  bisectors  of  the  angles  of  a  rectangle  B 
form  a  square. 

(By  Ex.  50,  EFGH  is  a  rectangle.    Now  prove 
AF  =  BH and  AE  =  BE.)  A 


85.  If  D  is  the  n)iddle  point  of  side  BC  of 
triangle  ABC,  and  BE  and  CF  are  perpendiculars 
from  B  and  C  to  AD,  produced  if  necessary,  prove 
BE  =  CF.  B 

86.  The  angle  at  the  vertex  of  isosceles  triangle 
ABC  is  equal  to  twice  the  sum  of  the  equal  angles 
B  and  C.  If  CD  be  drawn  perpendicular  to  BC, 
meeting  BA  produced  at  D,  prove  triangle  ACD 
equilateral. 

(Prove  each  Z  of  A  ACD  equal  to  60°.) 

87.  If  angle  B  of  triangle  ABC  is  greater  than  angle  C,  and  BD 
be  drawn  to  ^  C  making  AD  =  AB,  prove 

ZADB=l(B+  C),  and  ZCBD  =  l(B-  C). 
(Fig.  of  Prop.  XXXII.) 

88.  How  many  sides  are  there  in  the  polygon  the  sum  of  whose 
interior  angles  exceeds  the  sum  of  its  exterior  angles  by  540°  ? 


70 


PLANE   GEOMETRY.— BOOK  I. 


89.  The  sum  of  the  lines  drawn  from  any  point 
within  a  triangle  to  the  vertices  is  greater  than  the 
half-sum  of  the  three  sides. 

(Apply  §  61  to  each  of  the  ^ABD,  ACD,  and 
BCD.) 

90.  The  sum  of  the  lines  drawn  from  any  point  within  a  triangle 
to  the  vertices  is  less  than  the  sum  of  the  three  sides.     (§  48. ) 

(Fig.  of  Ex.  89.) 

91.  If  D,  E,  and  F  are  points  on  the  sides  A  B, 
BC,  and  CA,  respectively,  of  equilateral  triangle 
ABC,  such  that  AD  =  BE  =  CF,  prove  DEF  an 
equilateral  triangle. 

(Prove  AADF,  BDE,  and  C^i?' equal.) 

92.  If  E,  F,  G,  and  H  are  points  on  the 
sides  AB,  BC,  CD,  and  DA,  respectively,  of 
parallelogram  ABCD,  such  that  AE  =  CG  and 
BF  =  DH,  prove  EFGH  a  parallelogram. 


93.  If  E,  F,  G,  and  H  are  points  on  sides  AB, 
BC,  CD,  and  DA,  respectively,  of  square  ABCD, 
such  that  AE=  BF=  CG  =  DH,  prove  EFGH  a 
square. 

(First  prove  EFGH  equilateral.  Then  prove 
^FEH=^\) 

94.  If  on  the  diagonal  BD  of  square  ABCD  a 
distance  BE  be  taken  equal  to  AB,  and  EFhe  drawn 
perpendicular  to  BD,  meeting  AD  at  F,  prove  that 
AF=EF=  ED. 

95.  Prove  the  theorem  of  §  127  by  drawing  lines 
from  any  point  within  the  polygon  to  the  vertices. 
(§35.) 

96.  If  CD  is  the  perpendicular  from  the  ver- 
tex of  the  right  angle  to  the  hypotenuse  of  right 
triangle  ABC,  and  CE  the  bisector  of  angle  C, 
meeting  AB  at  E,  prove  ZDCE  equal  to  one-half 
the  difference  of  angles  A  and  B.  B 

(To  prove  Z  DCE  =  ^  (ZA  -  Z  B).) 

97.  State  and  prove  the  converse  of  Ex.  70,  p.  68. 
(Fig.  of  Prop.  XLIV.    Prove  the  sides  all  equal,) 


RECTILINEAR  FIGURES. 


71 


98.    State  and  prove  the  converse  of  Ex.  75,  p.  68. 
(Fig.  of  Ex.  78.     Prove  AACF  and  BDE  equal.) 


99.  D  is  any  point  in  base  BC  oi  isosceles 
triangle  ABC.  The  side  AG  \^  produced  from  C  to 
E,  so  that  CE  =  CD,  and  DE  is  drawn  meeting  AB 
at  F.    Prove  Z  AFE  =  3  Z  AEF. 

(Z  AFE  is  an  ext.  Z  of  A  BFD.) 

100.  If  ABC  and  ABD  are  two  triangles 
on  the  same  base  and  on  the  same  side  of  it, 
such  that  AC  =  BD  and  AD  =  BC,  and  AD 
and  BC  intersect  at  0,  prove  triangle  OAB 
isosceles. 

101.  If  D  is  the  middle  point  of  side  ^C  of  equi 
lateral  triangle  ABC,  and  DE  be  drawn  perpen 
dicular  to  BC,  prove  EC  =  ^  BC. 

(Draw  DF  to  the  middle  point  of  BC.) 

102.  If  in  parallelogram  A  BCD,  E  and  F 
are  the  middle  points  of  sides  BC  and  AD,  re- 
spectively, prove  that  lines  AE  and  CF  trisect 
diagonal  BD.  A 

(By  §  131,  AE  bisects  BH,  and  CF  bisects  DG.) 

103.  If  CD  is  the  perpendicular  from  C  to  the 
hypotenuse  of  right  triangle  ABC,  and  E  is  the 
middle  point  of  AB,  prove  ZDCE  equal  to 
the  difference  of  angles  A  and  5.     (Ex.  83.) 

104.  If  one  acute  angle  of  a  right  triangle  is  double  the  other,  the 
hypotenuse  is  double  the  shorter  leg. 

(Fig.  of  Ex.  86.    Draw  CA  to  middle  point  of  BD.) 

105.  If  AC  be  drawn  from  the  vertex  of  the  right  angle  to  the 
hypotenuse  of  right  triangle  BCD  so  as  to  make  ZACD  =  ZD,  it 
bisects  the  hypotenuse. 

(Fig.  of  Ex.  74.     Prove  A  ABC  isosceles.) 


106.    If  D  is  the  middle  point  of  side  BC  of 
triangle  ABC,  prove  AD>^iAB -\- AC  -  BC). 


Note.     For  additional  exercises  on  Book  I.,  see  p.  220. 


Book  II. 

THE    CIRCLE. 
DEFINITIONS. 

142.  A  circle  (O)  is  a  portion  of  a  plane  bounded  by  a 
curve  called  a  circiimference,  all  points 

of  which  are  equally  distant  from  a 
point  within,  called  the  centre;  as 
ABGD. 

An  arc  is  any  portion  of  the  circum- 
ference ;  as  AB. 

A  radius  is  a  straight  line  drawn 
from  the  centre  to  the  circumference ; 
as  OA. 

A  diameter  is  a  straight  line  drawn  through  the  centre, 
having  its  extremities  in  the  circumference ;  as  AG. 

143.  It  follows  from  the  definition  of  §  142  that 

All  radii  of  a  circle  are  equal. 

Also,  all  its  diameters  are  equal,  since  each  is  the  sum  of 
two  radii. 

144.  Two  circles  are  equal  when  their  radii  are  equal. 

For  they  can  evidently  be  applied  one  to  the  other  so  that 
their  circumferences  shall  coincide  throughout. 

145.  Conversely,  the  radii  of  equal  circles  are  equal. 

146.  A  semi-circumference  is  an  arc  equal  to  one-half  the 
circumference. 

A  qiiadrant  is  an  arc  equal  to  one-fourth  the  circumference. 

Concentric  circles  are  circles  having  the  same  centre. 
72 


THE  CIRCLE 


73 


147.  A  chord  is  a  straight  line  joining  the  extremities  of 
an  arc ;  as  AB. 

The  arc  is  said  to  be  subtended  by  its 
chord. 

Every  chord  subtends  two  arcs;  thus 
chord  AB  subtends  arcs  AMB  and 
ACDB. 

When  the  arc  subtended  by  a  chord  is 
spoken  of,  that  arc  which  is  less  than  a 
semi-circumference  is   understood,  unless   the  contrary  is 
specified. 

A  segment  of  a  circle  is  the  portion  included  between  an 
arc  and  its  chord ;  as  AMBN. 

A  semicircle  is  a  segment  equal  to  one-half  the  circle. 

A  sector  of  a  circle  is  the  portion  included  between  an  arc 
and  the  radii  drawn  to  its  extremities ;  as  OCD. 


148.  A  central  angle  is  an  angle  whose  vertex  is  at  the 
centre,  and  whose  sides  are  radii ;  as  AOC. 

An  inscribed  angle  is  an  angle  whose  ver- 
tex is  on  the  circumference,  and  whose 
sides  are  chords ;  as  ABC. 

An  angle  is  said  to  be  inscribed  in  a 
segment  when  its  vertex  is  on  the  arc  of 
the  segment,  and  its  sides  pass  through  the 
extremities  of  the  subtending  chord. 

Thus,  angle  B  is  inscribed  in  segment  ABC. 

149.  A  straight  line  is  said  to  be  tangent  to,  or  touch,  a 
circle  when  it  has  but  one  point  in  com- 
mon with  the  circumference ;  as  AB. 

In  such  a  case,  the  circle  is  said  to 
be  tangent  to  the  straight  line. 

The  common  point  is  called  the 
poi7it  of  contact,  or  poiyit  of  tangency. 

A  secant  is  a  straight  line  which 
intersects  the  circumference  in  two  points ;  as  CD. 


74 


PLANE  GEOMETRY. —BOOK  11. 


150.  Two  circles  are  said  to  be  tangent  to  each  other  when 
they  are  both  tangent  to  the  same  straight  line  at  the  same 
point. 

They  are  said  to  be  tangent  internally  or  externally  accord- 
ing as  one  circle  lies  entirely  within  or  entirely  without  the 
other. 

A  common  tangent  to  two  circles  is  a  straight  line  which 
is  tangent  to  both  of  them. 


151.  A  polygon  is  said  to  be  inscribed 
in  a  circle  when  all  its  vertices  lie  on  the 
circumference ;  as  ABCD. 

In  such  a  case,  the  circle  is  said  to  be 
circumscribed  about  the  polygon. 

A  polygon  is  said  to  be  inscriptible 
when  it  can  be  inscribed  in  a  circle. 

A  polygon  is  said  to  be  circumscribed 
about  a  circle  when  all  its  sides  are  tan- 
gent to  the  circle ;  as  EFGH. 

In  such  a  case,  the  circle  is  said  to  be 
itiscribed  in  the  polygon. 


Prop.  I.     Theorem. 

152.  Every  diameter  bisects  the  circle  and  its  circumference. 

B 


Given  AC  a  diameter  of  O  ABCD. 

To  Prove  that  AC  bisects  the  O,  and  its  circumference. 


THE   CIRCLE.  75 

Proof.  Superpose  segment  ABC  upon  segment  ADC,  by 
folding  it  over  about  AC  as  an  axis. 

Then,  arc  ABC  will  coincide  with  arc  ADC;  for  other- 
wise there  would  be  points  of  the  circumference  unequally 
distant  from  the  centre. 

Hence,  segments  ABC  and  ADC  coincide  throughout, 
and  are  equal. 

Therefore,  AC  bisects  the  O,  and  its  circumference. 

Prop.  II.     Theorem. 

153.  A  straight  line  cannot  intersect  a  circumference  at 
more  than  two  points. 

0 

/\ 

/  \\ 
/    \  \ 


MA  B  C     N 

Given  0  the  centre  of  a  O,  and  MJ^  any  str.  line. 

To  Prove  that  MN  cannot  intersect  the  circumference 
at  more  than  two  points. 

Proof.  If  possible,  let  JOT  intersect  the  circumference 
at  three  points,  A,  B,  and  C;  draw  radii  OA,  OB,  and  OC. 

Then,  OA=OB=  OC  (§  143) 

We  should  then  have  three  equal  str.  lines  drawn  from 
a  point  to  a  str.  line. 

But  this  is  impossible ;  for  it  follows  from  §  49  that  not 
more  than  two  equal  str.  lines  can  be  drawn  from  a  point  to 
a  str.  line. 

Hence,  MN  cannot  intersect  the  circumference  at  more 
than  two  points. 

Ex.  1.  What  is  the  locus  of  points  at  a  given  distance  from  a  given 
point? 


76 


PLANE   GEOMETRY.— BOOK  II. 


Prop.  III.     Theorem. 

154.   In  equal  circles,  or  in  the  same  circle,  equal  central 
angles  intercept  equal  arcs  on  the  circumference. 


Given  ACB  and  A'C'B'  equal  central  A  of  equal  (D  AMB 
and  A'M'B',  respectively. 

To  Prove  arc  AB  =  arc  A'B'. 

Proof.     Superpose  sector  ABC  upon  sector  A^B'O  in  such 
a  way  that  Z  C  shall  coincide  with  its  equal  Z  C. 

Now,  AC  =  A' a  and  BC  =  B'C.  (§  145) 

Whence,  point  A  will  fall  at  A',  and  point  B  at  B'. 

Then,  arc  AB  will  coincide  with  arc  A'B' ;  for  all  points 
of  either  are  equally  distant  from  the  centre. 
.'.  arc  AB  =  arc  A'B'. 


Prop.  IV.     Theorem. 

155.   (Converse  of  Prop.  III.)     In  equal  circles,  or  in  the 
same  circle,  equal  arcs  are  intercepted  by  equal  central  angles. 


Given  ACB  and  A'C'B'  central  A  of  equal  ©  AMB  and 
A'M'B',  respectively,  and  arc  AB  =  arc  A'B'. 


THE   CIRCLE. 


77 


To  Prove  AC  =  ^C'. 

Proof.  Since  the  (D  are  equal,  we  may  superpose  OAMB 
upon  O  A'M'B'  in  such  a  way  that  point  A  shall  fall  at  A', 
and  centre  C  at  O', 

Then  since  arc  AB  =  arc  A'B',  point  ^  will  fall  at  B'. 

Whence,  radii  AC  and  BC  will  coincide  with  radii  A'C 
and  5'C",  respectively.  (Ax.  3) 

Hence,  Z  G  will  coincide  with  Z  C. 
.-.  ZC'=ZC". 

156.  Sch.     In  equal  circles,  or  in  the  same  circle, 

1.  The  greater  of  two  central  angles  intercepts  the  greater 
arc  on  the  circumference. 

2.  TJie  greater  of  two  arcs  is  intercepted  by  the  greater  cen- 
tral angle. 

Prop.  V.     Theorem. 

157.  In  equal  circles,  or  in  the  same  circle,  equal  chords 
subtend  equal  arcs. 


M  M' 

Given,  in  equal  (D  AMB  and  A'M'B', 

chord  AB  =  chord  A'B'. 

To  Prove  arc  AB  =  arc  A'B'. 

Proof.     Draw  radii  AC,  BC,  A'C,  and  B'C. 

Then  in  A  ABC  and  A' B'C,  by  hyp., 
AB  =  A'B'. 

Also,  AC  =  A'C  and  BC  =  B'C. 

.-.  AABC=AA'B'C'. 

.'.  zc=zc. 

.'.  SLicAB  =  2iVcA'B'. 


(?) 

(?) 

(?) 

(§  154) 


78 


PLANE   GEOMETRY.  — BOOK  II. 


Prop.  VI.     Theorem. 

158.   (Converse  of  Prop.  V.)     In  equal  circles,  or  in  the 
same  circle,  equal  arcs  are  subtended  by  equal  chords. 

(Fig.  of  Prop.  V.) 

Given,  in  equal  ©  AMB  and  AM'B\  arc  AB  =  arc  AB' ; 
and  chords  AB  and  A'B\ 

To  Prove  chord  AB  =  chord  A'B'. 

(Prove  A  ABC  =  A  A'B'C,  by  §  63.) 


Ex.  2.  If  two  drcumferences  intersect  each  other,  the  distance 
between  their  centres  is  greater  than  the  difference  of  their  radii. 

(§  62.) 

Prop.  VII.     Theorem. 

159.  In  equal  circles,  or  in  the  same  circle,  the  greater  of 
two  arcs  is  subtended  by  the  greater  chord;  each  arc  being  less 
than  a  semircircumfereyice. 


Given,  in  equal  (D  AMB  and  A'M^B',  arc  AB  >  arc  A^B\ 
each  arc  being  <  a  semi-circumference,  and  chords  AB  and 

a:b'. 

To  Prove  chord  AB  >  chord  A'B'. 

Proof.     Draw  radii  AC,  BC,  A'C,  and  B'C. 
Then  in  A  ABC  and  A'B'C, 

AC  =  A'C  and  BG  =  B'C.  (?) 

And  since,  by  hyp.,  arc  AB  >  arc  A'B',  we  have 

ZC>ZC.  (§156,2) 

.-.  chord  AB  >  chord  A'B'.  (§  91) 


THE  CIRCLE. 


79 


Prop.  VIII.     Theorem. 

160.  (Converse  of  Prop.  VII.)  In  equal  circles,  or  in  the 
same  circle,  the  greater  of  two  chords  subtends  the  greater  arc; 
each  arc  being  less  than  a  semi-circumference. 

(Fig.  of  Prop.  VII.) 

(ZC>Z  C,  by  §  92 ;  the  theorem  follows  by  §  156,  1.) 

161.  Sch.  If  each  arc  is  greater  than  a  semi-circumfer- 
ence, the  greater  arc  is  subtended  by  the  less  chord ;  and 
conversely  the  greater  chord  subtends  the  less  arc. 

Prop.  IX.     Theorem. 

162.  TJie  diameter  perpendicular  to  a  chord  bisects  the 
chord  and  its  subtended  arcs. 


Given,  in  O  ABD,  diameter  CD  _L  chord  AB. 

To  Prove  that  CD  bisects  chord  AB,  and  arcs  ACB  and 
ADB. 

Proof.     Let  0  be  the  centre  of  the  O,  and  draw  radii  OA 
and  OB. 

Then,  OA  =  OB.  (?) 

Hence,  A  OAB  is  isosceles. 

Therefore,  CD  bisects  AB,  and  Z  AOB.  (§  94) 

Then  since  Z  AOC=  Z  BOC,  we  have 

Sivc.  AC  =  arc.  BC.  (§  154) 

Again,  Z  AOD  =  ZBOD.  (§  31,  2) 

.-.  SiVGAD=avGBD.  (?) 

Hence,  CD  bisects  AB,  and  arcs  ACB  and  ADB. 


80  PLANE   GEOMETRY.— BOOK  II. 

163.  Cor.  The  perpendicular  erected  at  the  middle  point 
of  a  chord  passes  through  the  centime  of  the  circle,  and  bisects 
the  arcs  subtended  by  the  chord. 

EXERCISES. 

3.  The  diameter  which  bisects  a  chord  is  perpendicular  to  it  and 
bisects  its  subtended  arcs.     (§43.) 

(Fig.  of  Prop.  IX.     Given  diameter  CD  bisecting  chord  AB.) 

4.  The  straight  line  which  bisects  a  chord  and 
its  subtended  arc  is  perpendicular  to  the  chord.  a, 

(By  §  158,  chord  AC  =  chord  BC.) 


Prop.  X.     Theorem. 

164.   In  the  same  circle,  or  in  equal  circles,  equal  chords  are 
equally  distant  from  the  centre. 


Given  AB  and  CD  equal  chords  of  O  ABC,  whose  centre 
is  0,  and  lines  OE  and  OF  ±  to  AB  and  CD,  respectively. 
To  Prove  OE  =  OF.  (§47) 

Proof.     Draw  radii  OA  and  0(7. 
Then  in  rt.  A  OAE  and  OCF, 

OA  =  OC  (?) 

Now,  E  is  the  middle  point  of  AB,  and  F  of  CD.    (§  162) 
.-.  AE=CF, 
being  halves  of  equal  chords  AB  and  CD,  respectively. 

.-.  A  OAE  =  A  OCF.  (?) 

.-.  OE=OF.  (?) 


THE   CIRCLE.  gj 

Prop.  XI.     Theorem. 

165.  (Converse  of  Prop.  X.)  In  the  same  circle,  or  in  equal 
circles,  chords  equally  distant  from  the  centre  are  equal. 

(Fig.  of  Prop.  X.) 

Given  0  the  centre  of  O  ABC,  and  AB  and  CD  chords 
equally  distant  from  0. 

To  Prove  chord  AB  =  chord  CD. 

(Rt.  A  OAE  =  rt.  A  OCF,  and  AE=CF,  E  is  the  middle 
point  of  AB,  and  F  of  CD.) 

Prop.  XII.     Theorem. 

166.  In  the  same  circle,  or  in  equal  circles,  the  less  of  two 
chords  is  at  the  greater  distance  from  the  centre. 

B 


Given,  in  O  ABC,  chord  AB  <  chord  CD,  and  Js  OF  and 
OG  drawn  from  centre  0  to  AB  and  CD,  respectively. 
To  Prove  OF>OQ. 

Proof.         Since  chord  AB  <  chord  CD,  we  have 

arc^S<arcCZ>.  (§  160) 

Lay  off  arc  CE  =  arc  AB,  and  draw  line  CE. 

.:  chord  CE  =  chord  AB.  (§  158) 

Draw  line  OHl.  CE,  intersecting  CD  at  K. 

.-.  0H=  OF.  (§  164) 

But,  OH>OK. 

And,  0K>00.  '  (?) 

Whence,  OH,  or  its  equal  OF,  is  >  OG. 


82  PLANE   GEOMETRY.— BOOK  11. 


Prop.  XIII.     Theorem. 

167.  (Converse  of  Prop.  XII.)  In  the  same  circle,  or  in 
equal  circles,  if  two  chords  are  unequally  distant  from  the 
centre,  the  more  remote  is  the  less. 


Given  0  the  centre  of  (d  ABC,  and  chord  AB  more  re- 
mote from  0  than  chord  CD. 

To  Prove  chord  AB  <  chord  CD. 

Proof.     Draw  lines  00.  and  QUI.  to  AB  and  CD  respec- 
tively, and  on  OC  lay  off  0K=  OH. 
Through  K  draw  chord  EFl.  OK. 

.'.  chord  EF=  chord  CD.  (§  165) 

Now,  chord  AB  II  chord  EF.  (§  54) 

Then  it  is  evident  that  arc  AB  is  <  arc  EF,  for  it  is  only 
a  portion  of  arc  EF. 

.:  chord  AB  <  chord  EF.  (§  159) 

.-.  chord  ^5  <  chord  (7i>. 

168.  Cor.  A  diameter  of  a  circle  is  greater  than  any 
other  chord;  for  a  chord  which  passes  through  the  centre  is 
greater  than  any  chord  which  does  not.  (§  167) 

EXERCISES. 


5.   The  diameter  which  bisects  an  arc  bisects  its 
chord  at  right  angles. 


6.   The  perpendiculars  to  the  sides  of  an  inscribed  quadrilateral 
at  their  middle  points  meet  in  a  common  point.     (§  163.) 


THE  CIRCLE. 


83 


Prop.  XIV.     Theorem. 

169.  A  straight  line  perpendicular  to  a  radius  of  a  circle 
at  its  extremity  is  tangent  to  the  circle. 


ADC  B 

Given  line  AB±  to  radius  OC  of  O  EC  at  G. 
To  Prove  AB  tangent  to  the  O. 

Proof.     Let  D  be  any  point  of  AB  except  C,  and  draw 
line  on. 

.'.  oD>oa  (?) 

Therefore,  point  D  lies  without  the  O. 
Then,  every  point  of  AB  except  C  lies  without  the  O, 
and  AB  is  tangent  to  the  O.  (§  149) 


Prop.  XV.     Theorem. 

170.   (Converse  of  Prop.  XIV.)     A  tangent  to  a  circle  is 
perpendicular  to  the  radius  drawn  to  the  point  of  contact. 


A  C  B 

Given  line  AB  tangent  to  O  EC  at  C,  and  radius  OC. 
To  Prove  OC±AB. 

(OC  is  the  shortest  line  that  can  be  drawn  from  0  to  AB.) 

171.   Cor.    A  line  perpendicular  to  a  tangent  at  its  point 
of  contact  passes  through  the  centre  of  the  circle. 


84 


PLANE   GEOMETRY.— BOOK  II. 


Prop.  XVI.     Theorem. 

172.    Two  parallels  intercept  equal  arcs  on  a  circumference. 
Case  I.    When  one  line  is  a  tangent  and  the  other  a  secant. 

^  B 


Given  AB  a  tangent  to  O  CED  at  E,  and  CD  a  secant  II 
AB,  intersecting  the  circumference  at  G  and  D. 
To  Prove  arc  CE  =  arc  DE. 

Proof.    Draw  diameter  EF. 

.'.  EF  LAB. 

.-.  EF  LCD. 

.'.  arc  CE  =  arc  DE. 


(§  170) 

(?) 
(§  162) 


Case  H.    When  both  lines  are  secants. 


Given,  in  O  ABC,  AB  and  CD  II  secants,  intersecting  the 
circumference  at  A  and  B,  and  C  and  D,  respectively. 

To  Prove  sltg  AC=  arc  BD. 

Proof.     Draw  tangent  EF  II  AB,  touching  the  O  at  G. 

.'.  EF  II  CD.  (?) 

Now,  arc  AG  =  arc  BG, 

and  arc  CG  =  arc  DG.  (§  172,  Case  I) 


THE   CIRCLE. 

Subtracting,  we  have 

arc  AG  —  arc  CG  =  arc  BG  —  arc  DG. 

.'.  arc  AC  =  arc  BD. 

Case  III.      When  both  lines  are  tangents. 

E 


85 


Given,  in  O  EGF,  AB  and  CD  II  tangents,  touching  the  O 
at  E  and  F,  respectively. 
To  Prove  arc  EGF=  arc  EHF. 

(Draw  secant  GH II  AB.) 

173.  Cor.    The  straight  line  joining  the  points  of  contact 
of  two  imrallel  tangents  is  a  diameter. 

Prop.  XVII.     Theorem. 

174.  Tlie  tangents  to  a  circle  from  an  outside  point  are 
equal. 


(Rt.  A  GAB  =  rt.  A  OAC,  by  §  90 ;  then  AB  =  AC.) 

175.   Cor.     From  equal  A  GAB  and  GAC, 

Z  GAB  =  Z  GAC  and  Z  AGB  =  Z  AGC 

Then,  the  line  joining  the  centre  of  a  circle  to  the  point  of 
intersection  of  two  tangents  makes  equal  angles  with  the  tan- 
gerits,  and  also  with  the  radii  drawn  to  the  points  of  contact. 


86 


PLANE   GEOMETRY.— BOOK  II. 


Prop.  XVIII.     Theorem. 

176.    Through  three  jjoints,  not  in  the  same  straight  line, 
a  drcumference  can  he  draivn,  and  but  one. 


Given  points  A,  B,  and  C,  not  in  the  same  straight  line. 

To  Prove  that  a  circumference  can  be  drawn  through  A, 
B,  and  C,  and  but  one. 

Proof.  Draw  lines  AB  and  BC,  and  lines  Z>i^ and  EG  ±  to 
AB  and  BC,  respectively,  at  their  middle  points,  meeting  at  0. 

Then  0  is  equally  distant  from  A,  B,  and  C.  (§  137) 

Hence,  a  circumference  described  with  0  as  a  centre  and 
OA  as  a  radius  will  pass  through  A,  B,  and  C. 

Again,  the  centre  of  any  circumference  drawn  through  A, 
B,  and  C  must  be  in  each  of  the  Js  DF  and  EG.  (§  42) 

Then  as  DF  and  EG  intersect  in  but  one  point,  only  one 
circumference  can  be  drawn  through  A,  B,  and  C. 

177.  Cor.  Two  cii-cumferences  can  intersect  in  hut  two 
'points;  for  if  they  had  three  common  points,  they  would 
have  the  same  centre,  and  coincide  throughout. 

Prop.  XIX.     Theorem. 

178.  If  two  circumferences  intersect,  the  straight  line  joining 
their  centres  bisects  their  common  chord  at  right  angles. 


THE   CIRCLE. 


87 


Given  0  and  0'  the  centres  of  two  (D,  whose  circumfer- 
ences intersect  at  A  and  B,  and  lines  00'  and  AB. 
To  Prove  that  00'  bisects  AB  at  rt.  A. 
(The  proposition  follows  by  §  43.) 


Prop.  XX.     Theorem. 

179.   If  two  circles  are  tangent  to  each  other,  the  straight 
line  joining  their  centres  passes  through  their  point  of  contact. 

B 


Given  0  and  0'  the  centres  of  two  CD,  which  are  tangent 
to  line  AB  at  A. 

To  Prove  that  str.  line  joining  0  and  0'  passes  through  A. 

(Draw  radii  OA  and  O'A ;  since  these  lines  are  ±  AB, 
OAO'  is  a  str.  line  by  §  37 ;  the  proposition  follows  by  Ax.  3.) 

EXERCISES. 


7.    The  straight  line  which  bisects  the  arcs  sub- 
tended by  a  chord  bisects  the  chord  at  right  angles. 


8.    The  tangents  to  a  circle  at  the  extremities  of  a  diameter  are 
parallel.  D 


9.   If  two  circles  are  concentric,  any  two  chords 
of  the  greater  which  are  tangent  to  the  less   are    C 
equal.     (§  165.) 


10.  The  straight  line  drawn  from  the  centre  of  a  circle  to  the  point 
of  intersection  of  two  tangents  bisects  at  right  angles  the  chord  joining 
their  points  of  contact.     (§  174.) 


88         PLANE  GEOMETKY.— BOOK  II. 

ON  MEASUREMENT. 

180.  The  ratio  of  a  magnitude  to  another  of  the  same 
kind  is  the  quotient  of  the  first  divided  by  the  second. 

Thus,  if  a  and  h  are  quantities  of  the  same  kind,  the  ratio 

of  a  to  6  is  -;  it  may  also  be  expressed  a  :  b. 
h 

A  magnitude  is  measured  by  finding  its  ratio  to  another 
magnitude  of  the  same  kind,  called  the  uyiit  of  measure. 

The  quotient,  if  it  can  be  obtained  exactly  as  an  integer 
or  fraction,  is  called  the  numerical  measure  of  the  magnitude. 

181.  Two  magnitudes  of  the  same  kind  are  said  to  be 
commensurable  when  a  unit  of  measure,  called  a  common 
measure,  is  contained  an  integral  number  of  times  in  each. 

Thus,  two  lines  whose  lengths  are  2|  and  3f  inches  are  commensu- 
rable ;  for  the  common  measure  i^  inch  is  contained  an  integral  num- 
ber of  times  in  each  ;  i.e.,  55  times  in  the  first  line,  and  76  times  in 
the  second. 

Two  magnitudes  of  the  same  kind  are  said  to  be  incommen- 
surable when  no  magnitude  of  the  same  kind  can  be  found 
which  is  contained  an  integral  number  of  times  in  each. 

For  example,  let  AB  and  CD  be  two  lines  such  that 

CD 

As  V2  can  only  be  obtained  approximately,  no  line,  however 
small,  can  be  found  which  is  contained  an  integral  number  of 
times  in  each  line,  and  AB  and  CD  are  incommensurable. 

182.  A  magnitude  which  is  incommensurable  with  respect 
to  the  unit  has,  strictly  speaking,  no  numerical  measure 
(§  180);  still  if  CD  is  the  unit  of  measure,  and  ^=  V2, 
we  shall  speak  of  V2  as  the  numerical  measure  of  AB. 

183.  It  is  evident  from  the  above  that  the  ratio  of  two 
magnitudes  of  the  same  kind,  whether  commensurable  or 
incommensurable,  is  equal  to  the  ratio  of  their  numerical 
measures  when  referred  to  a  common  unit. 


THE   CIRCLE. 


THE  METHOD  OF  LIMITS. 


184.  A  variable  quantity,  or  simply  a  variable,  is  a  quan- 
tity which  may  assume,  under  the  conditions  imposed  upon 
it,  an  indefinitely  great  number  of  different  values. 

185.  A  constant  is  a  quantity  which  remains  unchanged 
throughout  the  same  discussion. 

186.  A  limit  of  a  variable  is  a  constant  quantity,  the  dif- 
ference between  which  and  the  variable  may  be  made  less 
than  any  assigned  quantity,  however  small,  but  cannot  be 
made  equal  to  zero. 

In  other  words,  a  limit  of  a  variable  is  a  fixed  quantity 
to  which  the  variable  approaches  indefinitely  near,  but 
never  actually  reaches. 

187.  Suppose,  for  example,  that  a  point  moves  from  A 
towards  B  under  the  condition  that  it  ^  C      D  E  B 

shall  move,  during  successive  equal  in-    I 1 1 — I — I 

tervals  of  time,  first  from  A  to  C,  half-way  between  A  and 
B ;  then  to  D,  half-way  between  C  and  B ;  then  to  E,  half- 
way between  D  and  B]  and  so  on  indefinitely. 

In  this  case,  the  distance  between  the  moving  point  and 
B  can  be  made  less  than  any  assigned  distance,  however 
small,  but  cannot  be  made  equal  to  0. 

Hence,  the  distance  from  A  to  the  moving  point  is  a 
variable  which  approaches  the  constant  distance  AB  as  a 
limit. 

Again,  the  distance  from  the  moving  point  to  5  is  a 
variable  which  approaches  the  limit  0. 

As  another  illustration,  consider  the  series 
1     11     1    _i_    ... 

-*-'     2"'     4'    "8?     16'  J 

where  each  term  after  the  first  is  one-half  the  preceding. 

In  this  case,  by  taking  terms  enough,  the  last  term  may 
be  made  less  than  any  assigned  number,  however  small,  but 
cannot  be  made  actually  equal  to  0. 


90  PLANE   GEOMETKY.— BOOK  11. 

Then,  the  last  term  of  the  series  is  a  variable  which  ap- 
proaches the  limit  0  when  the  number  of  terms  is  indej&- 
nitely  increased. 

Again,  the  sum  of  the  first  two  terms  is  1^ ; 
the  sum  of  the  first  three  terms  is  If ; 
the  sum  of  the  first  four  terms  is  1 J ;  etc. 

In  this  case,  by  taking  terms  enough,  the  sum  of  the 
terms  may  be  made  to  differ  from  2  by  less  than  any  as- 
signed number,  however  small,  but  cannot  be  made  actually 
equal  to  2. 

Then,  the  sum  of  the  terms  of  the  series  is  a  variable 
which  approaches  the  limit  2  when  the  number  of  terms 
is  indefinitely  increased. 

188.  The  Theorem  of  Limits.  If  ttvo  variables  are  always 
equal,  and  each  approaches  a  limit,  the  limits  are  equal. 

AM  C        B 

\ \ I  I 

A'        M'  B' 

I \ I 

Given  AM  and  A'M^  two  variables,  which  are  always 
equal,  and  approach  the  limits  AB  and  A'B',  respectively. 

To  Prove  AB  =  A'B'. 

Proof.  If  possible,  let  AB  be  >  A'B';  and  lay  off,  on 
AB,AO=A'B'. 

Then,  variable  AM  may  have  values  >  AC,  while  vari- 
able A'M'  is  restricted  to  values  <  AC;  which  is  con- 
trary to  the  hypothesis  that  the  variables  are  always  equal. 

Hence,  AB  cannot  be  >  A'B'. 

In  like  manner,  it  may  be  proved  that  AB  cannot  be 
<A'B'. 

Therefore,  since  AB  can  be  neither  >,  nor  <,A'B',  we 
have 

AB  =  A'B'. 


THE   CIRCLE.  gj 

MEASUREMENT  OF  ANGLES. 

Prop.  XXI.     Theorem. 

189.   In  the  same  circle,  or  in  equal  circles,  two  central 
angles  are  in  the  same  ratio  as  their  intercepted  arcs. 

Case  I.    When  the  arcs  are  commensurable  (*§  181). 


Given,  in  QABC,  AOB  and  BOG  central  A  intercepting 
commensurable  arcs  AB  and  BO,  respectively. 

To  Prove  ZAOB^^tcAB^ 

Z  BOC      arc  BC 

Proof.  Since,  by  hyp.,  arcs  AB  and  BG  are  commensur- 
able, let  arc  AD  be  a  common  measure  of  arcs  AB  and  BC-, 
and  suppose  it  to  be  contained  4  times  in  arc  AB,  and  3 
times  in  arc  BG. 

SiTcAB  _  4  /-.  s 

''  arc 50 "3*  ^  ^ 

Drawing  radii  to  the  several  points  of  division  of  arc  AC, 
Z.AOB  will  be  divided  into  4  A,  and  A  BOG  into  3  A,  all 
of  which  A  are  equal.  (§  155) 

A  AOB     4 


*  A  BOG     3 

From  (1)  and  (2),  we  have 

AAOB^^x^AB 
A  BOG     slvcBG' 


(2) 
(?) 


92  PLANE  GEOMETRY.— BOOK  II. 

Case  n.      When  the  arcs  are  incommensurable  (§  181). 

B 


Given,  in  Q  ABG,AOB  and  BOC  central  A  intercepting 
incommensurable  arcs  AB  and  BC,  respectively. 

-,    -  ZAOB     SiTcAB 

To  Prove  .  ^^^,  = ^77- 

/.BOC      arcJ3C 

Proof.  Let  arc  AB  be  divided  into  any  number  of  equal 
arcs,  and  let  one  of  these  arcs  be  applied  to  arc^C  as  a 
unit  of  measure. 

Since  arcs  AB  and  BC  are  incommensurable,  a  certain 
number  of  the  equal  arcs  will  extend  from  B  to  O,  leaving 
a  remainder  C'C  less  than  one  of  the  equal  arcs. 

Draw  radius  OC. 

Then,  since  by  const.,  arcs  AB  and  BC  are  commensurable, 

ZbOC'^^^^W         (§  189,  Case  I.) 

Now  let  the  number  of  subdivisions  of  arc  AB  be  indefi- 
nitely increased. 

Then  the  unit  of  measure  will  be  indefinitely  diminished ; 
and  the  remainder  C'C,  being  always  less  than  the  unit,  will 
approach  the  limit  0. 

Then       Z  BOC  will  approach  the  limit  Z  BOC, 
and  arc  BC  will  approach  the  limit  arc  BC. 

Hence,    ^lA2^  will  approach  the  limit  ^^^^, 
'    ZBOC  ^^  ZBOC 

and  ^tgAB  ^.j^  approach  the  limit  ^:E^A^. 


THE  CIRCLE.  93 

^ow,  and — —  are  variables  which  are  always 

Z.B0O  SLTcBC       ZAOB  An 

equal,  and  approach  the  limits  — — ^  and  ^^'^  ^^  respec- 
tively. ^^^^         ^^^^^ 

By  the  Theorem  of  Limits,  these  limits  are  equal.   (§  188) 
.   Z  AOB  ^  arc  AB 
' '  Z  BOG     arc  BC' 

190.  Sch.  The  usual  unit  of  measure  for  arcs  is  the 
degree,  which  is  the  ninetieth  part  of  a  quadrant  (§  146). 

The  degree  of  arc  is  divided  into  sixty  equal  parts,  called 
minutes,  and  the  minute  into  sixty  equal  parts,  called 
seconds. 

If  the  sum  of  two  arcs  is  a  quadrant,  or  90°,  one  is  called 
the  complement  of  the  other ;  if  their  sum  is  a  semi-circum- 
ference, or  180°,  one  is  called  the  supplement  of  the  other. 

191.  Cor.  I.  By  §  154,  equal  central  A,  in  the  same  O, 
intercept  equal  arcs  on  the  circumference. 

Hence,  if  the  angular  magnitude  about  the  centre  of  a  O 
be  divided  into  four  equal  A,  each  Z  will  intercept  an  arc 
equal  to  one-fourth  of  the  circumference. 

That  is,  a  right  central  angle  intercepts  a  quadrant  on  the 
circumference.  (§  35) 

192.  Cor.  II.  By  §  189,  a  central  Z  of  n  degrees  bears 
the  same  ratio  to  a  rt.  central  Z  that  its  intercepted  arc 
bears  to  a  quadrant. 

But  a  central  Z  of  n  degrees  is  —  of  a  rt.  central  Z. 

Hence,  its  intercepted  arc  is  —  of  a  quadrant,  or  an  arc  of 
n  degrees. 

The  above  principle  is  usually  expressed  as  follows : 

A  ceiitral  angle  is  measured  by  its  intercepted  arc. 

This  means  simply  that  the  number  of  angular  degrees  in 
a  central  angle  is  equal  to  the  number  of  degrees  of  arc  in 
its  intercepted  arc. 


PLANE   GEOMETRY.  — BOOK  II. 


Prop.  XXII.     Theorem. 

193.  An  inscribed  angle  is  measured  by  one-half  its  inter- 
cepted arc. 

Case  I.     When  one  side  of  the  angle  is  a  diameter. 

A 


Oiven  AC  a  diameter,  and  AB  a  chord,  of  O  ABO. 
To  Prove  that  Z  BAG  is  measured  by  ^  arc  BG. 
Proof.     Draw  radius  OB ;  then,  OA  =  OB. 
Then  A  OAB  is  isosceles,  and  Z  B  =  ZA. 
But  since  BOG  is  an  ext.  Z  of  A  OAB, 
Z  BOG  =  ZA-\-ZB. 
.'.  ZB0G=2ZA,  or  ZA  =  iZBOG. 
But,  Z  BOG  is  measured  by  arc  BG. 
Whence,  ZAis  measured  by  ^  arc  BG. 

Case  11.     When  the  centre  is  within  the  angle. 
A 


(?) 
(?) 

(§  85,  1) 
(§  192) 


Given  AB  and  AG  chords  of  O  ABC,  and  the  centre  of 
the  O  within  Z  BAG. 

To  Prove  that  Z  BAG  is  measured  by  J  arc  BG. 


THE   CIRCLE. 


95 


Proof.   Draw  diameter  AD. 
Then,         Z  BAD  is  measured  by  |  arc  BD, 
and  Z  CAD  is  measured  by  ^  arc  CD.  (§  193,  Case  I) 

.-.  Z  BAD  +  Z  C^Z)  is  measured  by  |  arc  BD  +  ^  arc  0/>. 
.-.  Z.^^O  is  measured  by  ^  arc  ^0. 
Case  III.    When  the  centre  is  without  the  angle. 

A 


(The  proof  is  left  to  the  pupil.) 

194.  Cor.  I.  Angles  inscribed  in  the 
same  segment  are  equal. 

Given  A,  B,  and  (7  zi  inscribed  in  seg- 
ment ADE  of  O  ABC. 

To  Prove    ZA=Z.B  =  ZG. 
(The  proposition  follows  by  §  193.) 

195.  Cor.  II.  An  angle  inscribed  in  a 
semicircle  is  a  right  angle. 

Given  BC  a  diameter,  and  AB  and  AC  B 
chords,  of  O  ABD. 

To  Prove    Z  BAC  a  rt.  Z. 

Proof.  ZBAC  is  measured  by  ^  of 
180°,  or  90°.  (§  193) 

196.  Cor.  III.  Tlie  opposite  angles  of 
an  inscribed  quadrilateral  are  supplement- 
ary. 

For  their  sum  is  measured  by  ^  of  360°, 
or  180°.  (?) 


96 


PLANE   GEOMETRY.  — BOOK  II. 


Prop.  XXIII.     Theorem. 

197.    The  angle  between  a  tangent  and  a  chord  is  measured 
by  one-half  its  intercepted  arc. 

B 


Given  AE  a  tangent  to  O  BCD  at  B,  and  BQ  a  chord. 
To  Prove  that  Z  ABC  is  measured  by  i  arc  BO. 
Proof.     Draw  diameter  BD ;  then,  BD  A.  AE.  (?) 

Now  a  rt.  Z  is  measured  by  one-half  a  semi-circumference. 

.-.  Z  ABD  is  measured  by  ^  arc  BCD. 
Also,  Z  (7jBZ)  is  measured  by  \  arc  CD.  (§  193) 

.-.  Z  ABD  -  Z  (75i>  is  measured  by  |  arc  5(7i)  -|  arc  Oi>. 

.-.  Z  ABC  is  measured  by  ^  arc  BG. 
Similarly,    Z  ^JB(7  is  measured  by  ^  arc  BDG. 

Prop.  XXIV.     Theorem. 

198.  The  angle  between  two  chords,  intersecting  within  the 
circumference,  is  measured  by  one-half  the  sum  of  its  inter- 
cepted arc,  and  the  arc  intercepted  by  its  vertical  angle. 

-4. 


Given,  in  0  ABC,  chords  AB  and  CD  intersecting  within 
the  circumference  at  E. 


THE  CIRCLE. 


97 


To  Prove  that 

Z.  AEG  is  measured  by  \  (arc  ^(7  +  arc  BTf), 
Proof.     Draw  chord  BG. 
Then,  since  AEG  is  an  ext.  Z  of  A  BGE^ , 

ZAEG  =  ZB  +  ZG.  (?) 

But,  Z  5  is  measured  by  \  arc  AG, 

and  Z  C  is  measured  by  \  arc  5Z).  (?) 

.*.  Z  AE;(7  is  measured  by  ^  (arc  ^C-f-  arc  ^D). 

Prop.  XXV.    Theorem. 

199.  The  angle  between  tivo  secants,  intersecting  without 
the  circumference,  is  measured  by  one-half  the  difference  of 
the  intercepted  arcs. 


Given,  in  (DABG,  secants  AE  and  GE  intersecting  with- 
out the  circumference  at  E,  and  intersecting  the  circumfer- 
ence at  A  and  B,  and  G  and  D,  respectively. 

To  Prove  that  Z  ^  is  measured  by  ^  (arc  J. (7—  arc  BD). 

Proof.     Draw  chord  BG. 

Then  since  ABG  is  an  ext.  Z  of  A  BGE, 

ZABG=ZE  +  ZG.  (?) 

.-.  ZE  =  ZABG-ZG. 
But,        Z  ABG  is  measured  by  ^  arc  AG, 
and  ZG  is  measured  by  ^  arc  BD.  (?) 

.-.  Z  ^  is  measured  by  J  (arc  AG  — sue  BD). 


98 


PLANE    GEOMETRY.— BOOK   II. 


200.  Cor.  (Converse  of  §  196.)  If  the 
opposite  angles  of  a  quadrilateral  are  supple- 
mentary, the  quadrilateral  can  he  inscribed  in 
a  circle. 

Given,  in  quadrilateral  ABCD,  Z.A  sup.  Ji 
to  Z  C,  and  ZB  to  Z  Z> ;  also,  a  circumfer- 
ence drawn  through  A,  B,  and  G.       (§  176) 

To  Prove  that  D  lies  on  the  circumference. 

Proof.  Since  ZD  is  sup.  to  Z  B,  it  is  measured  by  ^  arc 
ABO.  (§  193) 

Then,  D  must  lie  on  the  circumference;  for  if  it  were 
within  the  O,  ZD  would  be  measured  by  ^  an  arc  >  ABC', 
and  if  it  were  without  the  O,  ZD  would  be  measured  by 
1  an  arc  <  ABC.  (§§  198,  199) 


Pkop.  XXVI.     Theorem. 

201.  The  angle  between  a  secant  and  a  tangent,  or  ttvo 
tangents,  is  measured  by  one-half  the  difference  of  the  inter- 
cepted arcs. 


Fig.  1. 


Fig.  2. 


1.  Given  AE  a  tangent  to  0  BDC  at  B,  and  EC  a  secant 
intersecting  the  circumference  at  C  and  D.     (Fig.  1.) 

To  Prove  that  Z^  is  measured  by  i  (arc  BFC—  arc  BD). 
(We  have  ZE  =  Z  ABC  -  Z  C.) 

2.  (In  Mg.  2,  Z  ^  =  Z  ABD  -  Z  BDE ;  then  use  §  197.) 


THE  CIRCLE.  99 

202.  Cor.    Since  a  circumference  is  an  arc  of  360°,  we 
have 
i  (arc  BFD  -  arc  BGD) 

=  i  (360°  -  arc  BGD  -  arc  BGD) 
=  i  (360°  -  2  arc  BGD) 
=  180°  -  arc  BGD. 
Then,  Z  E  is  measured  by  180°  —  arc  BGD. 
Hence,  the  angle  between  two  tangents  is  measured  by  the 
supplement  of  the  smaller  of  the  two  intercepted  arcs. 

EXERCISES. 

11.  If,  in  figure  of  §  197,  arc  BC  =  107°,  how  many  degrees  are 
there  in  angles  ABC  and  EBC? 

12.  If,  in  figure  of  §  198,  arc  AC  =  'I^%  and  ZAEC  =  5r,  how 
many  degrees  are  there  in  arc  BD  ? 

13.  If,  in  figure  of  §  199,  arc  ^C  =  117°,  and  ZC=  14°,  how  many 
degrees  are  there  in  angle  E? 

14.  If,  in  figure  of  §  199,  ^C  is  a  quadrant,  and  ZE  =  39°,  how 
many  degrees  are  there  in  arc  BD  ? 

15.  If,  in  Fig.  1  of  §  201,  arc  BFC  =  197°,  and  arc  CD  =  75°,  how 
many  degrees  are  there  in  angle  E  ? 

16.  If,  in  Fig.  1  of  §  201,  Z  J5:  =  53°,  and  arc  BD  is  one-fifth  of  the 
circumference,  how  many  degrees  are  there  in  arc  BFC  ? 

17.  If,  in  Fig.  2  of  §  201,  arc  BFD  is  thirteen-sixteenths  of  the 
circumference,  how  many  degrees  are  there  in  angle  E  ? 

18.  Three  consecutive  sides  of  an  inscribed  quadrilateral  subtend 
arcs  of  82°,  99°,  and  67°  respectively.  Find  each  angle  of  the  quad- 
rilateral in  degrees,  and  the  angle  between  its  diagonals. 

19.  Prove  Prop.  XXIV.  by  drawing  through  B  a  chord  parallel  to 
CD.     (§  172.) 

20.  Prove  Prop.  XXV.  by  drawing  through  B  a  chord  parallel 
to  CD. 

21.  Prove  Prop.  XXVI.  for  Fig.  1  by  drawing  through  D  a  chord 
parallel  to  AE. 

22.  An  angle  inscribed  in  a  segment  greater  than  a  semicircle  is 
acute  ;  and  an  angle  inscribed  in  a  segment  less  than  a  semicircle  is 
obtuse.     (§  193.) 


100 


PLANE   GEOMETRY.  — BOOK  II. 


23.  In  an  inscribed  trapezoid  the  non-parallel  sides  are  equal,  and 
also  the  diagonals. 

(The  non-parallel  sides,  and  also  the  diagonals,  subtend  equal  arcs.) 

24.  If  the  inscribed  and  circumscribed  circles  of  a  triangle  are  con- 
centric, prove  the  triangle  equilateral.     (§  165.) 

25.  If  AB  and  AC  are  the  tangents  from  point  A  to  the  circle 
whose  centre  is  0,  prove  Z  BAG  =  2  Z  OBC.     (Ex.  10,  p.  87.) 

c 

26.  If  two  chords  intersect  at  right  angles  within 
the  circumference  of  a  circle,  the  sum  of  the  oppo-      /  \      \  B 
site  intercepted  arcs  is  equal  to  a  semi-circumfer- 
ence. 


27.  Two  intersecting  chords  which  make 
equal  angles  with  the  diameter  passing  through 
their  point  of  intersection  are  equal.     (§  165.) 

(Prove  that  01f  =  0ir.) 


28.   Prove  Prop.  XXIII.  by  drawing  a  radius 
perpendicular  to  BO.     (§  162.) 


29.  If  AB  and  AC  are  two  chords  of  a  circle  making  equal  angles 
with  the  tangent  at  J.,  prove  AB  =  AC. 

30.  From  a  given  point  within  a  circle  and  not 
coincident  with  the  centre,  not  more  than  two  equal 
straight  lines  can  be  drawn  to  the  circumference. 

(If  possible,  let  AB,  AC,  and  AD  be  three  equal 
straight  lines  from  point  A  to  circumference  BCD; 
then,  by  §  163,  A  must  coincide  with  the  centre.) 

31.  The  sum  of  two  opposite  sides  of  a  circum- 
scribed quadrilateral  is  equal  to  the  sum  of  the  other 
two  sides.     (§  174.) 

(To  prove  AB  +  CD  =  AD  +  BC.) 


THE  CIRCLE.  .  IQl- 

32.  Prove  Prop,  VI.  by  superposition.  ,     >  J . ,  ^  ; '   - ,  •>  '.  '*,  i 

33.  In  a  circumscribed  trapezoid,  the  straight  line  joining  the 
middle  points  of  the  non-parallel  sides  is  equal  to  one-fourth  the 
perimeter  of  the  trapezoid.     (§  132.) 

34.  If  the  opposite  sides  of  a  circumscribed  quadrilateral  are  paral- 
lel, the  figure  is  a  rhombus  or  a  square.     (Ex.  31.) 

(Prove  the  sides  all  equal.) 

35.  If  tangents  be  drawn  to  a  circle  at  the  extremities  of  any  pair 
of  diameters  which  are  not  perpendicular  to  each  other,  the  figure 
formed  is  a  rhombus.     (Ex.  34.) 

36.  If  the  angles  of  a  circumscribed  quadrilateral  are  right  angles, 
the  figure  is  a  square. 

37.  If  two  circles  are  tangent  to  each  other  at  point  A,  the  tangents 
to  them  from  any  point  in  the  common  tangent  which  passes  through 
A  are  equal.     (§  174.) 

38.  If  two  circles  are  tangent  to  each  other 
externally  at  point  A,  the  common  tangent  which 
passes  through  A  bisects  the  other  two  common 
tangents.  (Ex.  37.) 

(To  prove  that  FG  bisects  BC  and  DE.) 

39.  The  bisector  of  the  angle  between  two  tangents  to  a  circle 
passes  through  the  centre. 

(The  bisector  of  the  Z  between  the  tangents  bisects  at  rt.  A  the 
chord  joining  their  points  of  contact.) 

40.  The  bisectors  of  the  angles  of  a  circumscribed  quadrilateral 
pass  through  a  common  point. 

41.  If  AB  is  one  of  the  non-parallel  sides  of  a  trapezoid  circum- 
scribed about  a  circle  whose  centre  is  0,  prove  AOB  a  right  angle. 
(§  175.) 

B 

42.  If  two  circles  are  tangent  to  each  other 
internally,  the  distance  between  their  centres  is 
equal  to  the  difference  of  their  radii. 

43.  Prove  the  theorem  of  §  168  by  drawing  radii  to  the  extremities 
of  the  chord.     (Ax.  4.) 

44.  Prove  the  theorem  of  §  202  by  drawing  radii  to  the  points  of 
contact  of  the  tangents.     (§  192.) 

45.  If  any  number  of  angles  are  inscribed  in  the  same  segment, 
their  bisectors  pass  through  a  common  point.     (§  193.) 


102 


PLANE   GEOMETRY.  — BOOK  II. 


.'  '  46.   Prove-  Prap.  XIII.  by  Beductio  ad  Absurdum.  (§§  164,  166.) 

47.   Two  chords  perpendicular  to  a  third  chord  at  its  extremities 
are  equal.     (§  158.) 


48.   If  two  opposite  sides  of  an  inscribed  quadrilat- 
eral are  equal  and  parallel,  the  figure  is  a  rectangle. 
(Arc  BCD  is  a  semi-circumference.) 


49.   If  the  diagonals  of  an  inscribed  quadrilateral  intersect  at  the 
centre  of  the  circle,  the  figure  is  a  rectangle.     (§  195.) 


50.  The  circle  described  on  one  of  the  equal 
sides  of  an  isosceles  triangle  as  a  diameter,  bisects 
the  base.    (§  195.) 


51.  If  a  tangent  be  drawn  to  a  circle  at  the  ex- 
tremity of  a  chord,  the  middle  point  of  the  sub- 
tended arc  is  equally  distant  from  the  chord  and 
from  the  tangent. 

(JB2>  bisects  Z^^C.) 


52.  If  sides  AB,  BC,  and  CD  of  an  inscribed  quadrilateral  sub- 
tend arcs  of  99°,  106°,  and  78°,  respectively,  and  sides  BA  and  CD 
produced  meet  at  E,  and  sides  AD  and  BC  Sit  F,  find  the  number  of 
degrees  in  angles  AED  and  AFB. 

53.  If  0  is  the  centre  of  the  circumscribed  circle  of  triangle  ABC, 
and  OD  be  drawn  perpendicular  to  BC,  prove 

ZBOD  =  ZA.     (§192.) 

54.  If  Z>,  E,  and  F  are  the  points  of  contact  of  sides  AB,  BC,  and 
CA  respectively  of  a  triangle  circumscribed  about  a  circle,  prove 

Z  DEF  =  90°  -  I  A.     (§  202.) 

55.  If  sides  AB  and  BC  of  an  inscribed  quadrilateral  ABCD  sub- 
tend arcs  of  69°  and  112°,  respectively,  and  angle  AED  between  the 
diagonals  is  87°,  how  many  degrees  are  there  in  each  angle  of  the 
quadrilateral  ? 

56.  If  any  number  of  parallel  chords  be  drawn  in  a  circle,  their 
middle  points  lie  in  the  same  straight  line.     (§  162.) 

57.  What  is  the  locus  of  the  middle  points  of  a  system  of  parallel 
chords  in  a  circle  ? 


THE  CIRCLE. 


103 


58.  What  is  the  locus  of  the  middle  points  of  a  system  of  chords 
of  given  length  in  a  circle  ? 

59.  If  two  circles  are  tangent  to  each  other, 
any  straight  line  drawn  through  their  point  of 
contact  subtends  arcs  of  the  same  number  of 
degrees  on  their  circumferences.     (§  197.) 

(Let  the  pupil  draw  the  figure  for  the  case 
when  the  (D  are  tangent  internally.) 


60.  If  a  straight  line  be  drawn  through  the 
point  of  contact  of  two  circles  which  are  tan- 
gent to  each  other  externally,  terminating  in 
their  circumferences,  the  radii  drawn  to  its 
extremities  are  parallel.    (§  73.) 

(Let  the  pupil  state  the  corresponding  theo- 
rem for  the  case  when  the  (D  are  tangent  internally.) 

61.  If  a  straight  line  be  drawn  through  the  point  of  contact  of  two 
circles  which  are  tangent  to  each  other  externally,  terminating  in  their 
circumferences,  the  tangents  at  its  extremities  are  parallel.     (§  73.) 

(Let  the  pupil  state  the  corresponding  theorem  for  the  case  when 
the  ©  are  tangent  internally.) 

62.  If  sides  AB  and  DC  of  inscribed  quadrilateral  ABCD  be  pro- 
duced to  meet  at  E,  prove  that  triangles  ACE  and  BDE^  and  also 
triangles  ADE  and  BCE,  are  mutually  equiangular. 

(For  second  part,  see  §  196.) 

63.  The  sum  of  the  angles  subtended  at  the 
centre  of  a  circle  by  two  opposite  sides  of  a  circum- 
scribed quadrilateral  is  equal  to  two  right  angles. 
(§  175.) 

(To  prove  Z  AOB  +  Z  COD  =  180°.) 


64.  If  a  circle  is  inscribed  in  a  right  triangle, 
the  sum  of  its  diameter  and  the  hypotenuse  is  equal 
to  the  sum  of  the  legs.     (§  174.) 


65.  If  a  circle  be  described  on  the  radius  of  another  circle  as  a 
diameter,  any  chord  of  the  greater  passing  through  the  point  of  con- 
tact of  the  circles  is  bisected  by  the  circumference  of  the  smaller. 
(§  196.) 


104 


PLANE   GEOMETRY. —BOOK  II. 


66.  If  sides  AB  and  CD  of  inscribed  quad- 
rilateral ABCD  make  equal  angles  with  the 
diameter  passing  through  their  point  of  inter- 
section, prove  AB  =  CD.     (§  165.) 


67.  If  angles^,  B,  and  G  of  circumscribed  quadrilateral  ylBC2> 
are  128°,  67°,  and  112°,  respectively,  and  sides  AB,  BC,  CD,  and  DA 
are  tangent  to  the  circle  at  points  E,  F,  6?,  and  H,  respectively,  find 
the  number  of  degrees  in  each  angle  of  quadrilateral  EFGH. 

68.  The  chord  drawn  through  a  given  point  within 
a  circle,  perpendicular  to  the  diameter  passing  through 
the  point,  is  the  least  chord  which  can  be  drawn 
through  the  given  point.     (§  165.) 

(Given  chords  AB  and  CD  drawn  through  P,  and 
AB  ±  OP.     To  prove  AB  <  CD.) 

69.  If  ADB,  BEC,  and  CFA  are  angles  inscribed  in  segments 
ABD,  BCE,  and  ACF,  respectively,  exterior  to  inscribed  triangle 
ABC,  prove  their  sum  equal  to  four  right  angles.     (§  196.) 

Note.    For  additional  exercises  on  Book  II. ,  see  p.  222. 


CONSTRUCTIONS. 

Prop.  XXVII.     Problem. 

203.  At  a  given  point  in  a  straight  line  to  erect  a  perpen- 
dicular to  that  line. 

First  Method. 


F,,- 


A    D  C  E     B 

Given  C  any  point  in  line  AB. 


THE  CIRCLE.  105 

Required  to  draw  a  line  ±  to  AB  at  C. 

Construction.  Take  points  D  and  E  on  AB  equally  dis- 
tant from  0. 

With  D  and  E  as  centres,  and  with  equal  radii,  describe 
arcs  intersecting  at  F^  and  draw  line  CF. 

Then,  CF  is  J_  to  AB  at  C. 

Proof.  By  cons.,  C  and  F  are  each  equally  distant  from 
D  and  E. 

Whence,  CF  is  ±  to  i)^  at  its  middle  point.  (?) 

Second  Method. 

E 
I 


Given  (7  any  point  in  line  AB. 

Required  to  draw  a  line  ±  to  ^B  at  C. 

Construction.  With  any  point  0  without  line  AB  as  a 
centre,  and  distance  OC  as  a  radius,  describe  a  circumfer- 
ence intersecting  AB  at  C  and  B. 

Draw  diameter  Z>^,  and  line  CE. 

Then,  O^  is  ±  to  ^B  at  C. 

Proof.  Z  i)(7^,  being  inscribed  in  a  semicircle,  is  a 
rt.  Z.  (§  195) 

.-.  CE  ±  CZ). 

Note.  The  second  method  of  construction  is  preferable  when  the 
given  point  is  near  the  end  of  the  line. 

EXERCISES. 

70.  Given  the  base  and  altitude  of  an  isosceles  triangle,  to  con- 
struct the  triangle. 

71.  Given  an  acute  angle,  to  construct  its  complement. 


106  PLANE  GEOMETRY.— BOOK  II. 

Prop.  XXVIII.     Problem. 

204.  From  a  given  point  without  a  straight  line  to  draw  a 
perpendicular  to  that  line. 


C 
/ 

f 


Given  C  any  point  without  line  AB. 

Required  to  draw  from  C  a  line  ±  to  AB. 

Construction.  With  (7  as  a  centre,  and  any  convenient 
radius,  describe  an  arc  intersecting  AB  2X  D  and  E. 

With  D  and  E  as  centres,  and  with  equal  radii,  describe 
arcs  intersecting  at  F. 

Draw  line  CF. 

Then,  CF  ±  AB. 

Proof.  Since,  by  cons.,  C  and  F  are  each  equally  distant 
from  D  and  F,  CF  is  ±  to  BF  at  its  middle  point.  (?) 

Prop.  XXIX.     Problem. 
205.    To  bisect  a  given  straight  line. 

"M 


\E 


-B 


% 


Given  line  AB. 


THE  CIRCLE.  IQ7 

Required  to  bisect  AB. 

Construction.     With  A  and  B  as  centres,  and  with  equal 
radii,  describe  arcs  intersecting  at  C  and  D. 
Draw  line  CD  intersecting  AB  at  E. 
Then,  E  is  the  middle  point  of  AB. 
(The  proof  is  left  to  the  pupil.) 

Prop.  XXX.     Problem. 
206.    To  bisect  a  given  arc. 


Given  arc  AB. 
Required  to  bisect  arc  AB. 

Construction.     With  A  and  B  as  centres,  and  with  equal 
radii,  describe  arcs  intersecting  at  C  and  D. 
Draw  line  CD  intersecting  arc  AB  at  E. 
Then  E  is  the  middle  point  of  arc  AB. 
Proof.     Draw  chord  AB. 

Then,  CD  is  ±  to  chord  AB  at  its  middle  point.  (?) 

Whence,  CD  bisects  arc  AB.  (§  163) 

EXERCISES. 

72.   Given  an  angle,  to  construct  its  supplement. 
•     73.    Given  a  side  of  an  equilateral  triangle,  to  construct  the  tri- 


74.  To  construct  an  angle  of  60^  (Ex.  73);  of  30°  (Ex.  71). 

75.  To  construct  an  angle  of  120°  (Ex.  72);  of  160°. 


108  PLANE   GEOMETRY.— BOOK  n. 

Prop.  XXXI.     Problem. 
207.   To  bisect  a  given  angle. 


Given  Z  AOB. 

Required  to  bisect  Z.  AOB. 

Construction.     With  0  as  a  centre,  and  any  convenient 
radius,  describe  an  arc  intersecting  OA  at  C,  and  OB  at  D. 

With  C  and  D  as  centres,  and  with  the  same  radius  as 
before,  describe  arcs  intersecting  at  E,  and  draw  line  OE. 

Then,  OE  bisects  Z  AOB. 

Proof.     Let  OE  intersect  arc  CD  at  F. 

By  cons.,  0  and  E  are  each  equally  distant  from  C  and  D. 

Whence,  OE  bisects  arc  CD  at  F  (§  206). 

.-.  ZCOF=ZDOF.  (?) 

That  is,  OE  bisects  Z  AOB. 

Prop.  XXXII.     Problem. 

208.   With  a  given  vertex  and  a  given  side,  to  construct  an 
angle  equal  to  a  given  angle. 


C  ^  A 

Given  0  the  vertex,  and  OA  a  side,  of  an  Z,  and  Z  0', 


THE  CIRCLE.  IQg 

Required  to  construct,  with  0  as  the  vertex  and  OA  as  a 
side,  an  Z  equal  to  Z  0'. 

Construction.  With  0'  as  a  centre,  and  any  convenient 
radius,  describe  an  arc  intersecting  the  sides  of  Z  0'  at  0 
and  Z> ;  and  draw  chord  CD. 

With  0  as  a  centre,  and  with  the  same  radius  as  before, 
describe  the  indefinite  arc  AE. 

With  J.  as  a  centre  and  CD  as  a  radius,  describe  an  arc 
intersecting  arc  AE  at  B,  and  draw  line  OB. 

Then,  ZAOB  =  ZO'. 

(The  chords  of  arcs  AB  and  CD  are  equal,  and  the  propo- 
sition follows  by  §§  157  and  155.) 

Prop.  XXXIII.     Problem. 

209.  Through  a  given  point  without  a  given  straight  li7ie,  to 
draw  a  parallel  to  the  line. 

/F 


/c 


-D 


/E 


Given  C  any  point  without  line  AB. 
Required  to  draw  through  C  a  line  ||  to  AB. 
Construction.     Through   C  draw  any  line   EF,  meeting 
AB  at  E,  and  construct  Z  FCD  =  Z  CEB.  (§  208) 

Then,  CD  ||  AB.  (?) 

EXERCISES. 

76.  To  construct  an  angle  of  45°  ;  of  135°  ;  of  22^°  ;  of  67^°. 

77.  Through  a  given  point  without  a  straight  line  to  draw  a  line 
making  a  given  angle  with  that  line. 

(Draw  through  the  given  point  a  II  to  the  given  line.) 


110 


PLANE   GEOMETRY.  — BOOK  U. 


Prop.  XXXIV.     Problem. 
210.    Given  two  angles  of  a  triangle,  to  find  the  third. 


<F 


-B 


Given  A  and  B  two  A  of  a,  A. 
Required  to  construct  the  third  Z. 

Construction.     At  any  point  E  of  the  indefinite  line  CD, 
construct  Z  DEF=  Z  A.  (§  208) 

Also,  Z  FEG  adjacent  to  Z  DEF,  and  equal  to  Z  B. 
Then,  Z  CEG  is  the  required  Z. 
(The  proof  is  left  to  the  pupil.) 


Prop.  XXXV.    Problem. 

211.   Given  tvjo  sides  and  the  included  angle  of  a  triangle, 
to  construct  the  triangle. 

ca 

m / 

ny 


Given  m  and  n  two  sides  of  a  A,  and  A^  their  included  Z. 

Required  to  construct  the  A. 

Construction.     Draw  line  AB  ==  m. 

Construct  Z  BAD  =  Z  A'.  (§  208) 

On  AD  take  AO  =  n,  and  draw  line  BC. 

Then,  ABC  is  the  required  A. 

212.   Sch.     The   problem  is  possible  for  any  values  of 
the  given  parts. 


THE  CIRCLE. 


Ill 


Prop.  XXXVI.     Problem. 

213.    Given  a  side  and  two  adjacent  angles  of  a  triangle^  to 
construct  the  triangle. 

D 


Given  a  side  m,  and  the  adj.  A  A'  and  B'  of  a  A. 
(The  construction  is  left  to  the  pupil.) 

214.  Sch.  I.  If  a  side  and  any  two  angles  of  a  triangle 
are  given,  the  third  angle  may  be  found  by  §  210,  and  the 
triangle  may  then  be  constructed  as  in  §  213. 

215.  Sch.  n.  The  problem  is  impossible  when  the  sum 
of  the  given  angles  is  equal  to,  or  greater  than,  two  right 
angles.  (§  84) 

Prop.  XXXVII.     Problem. 

216.  Given  the  three  sides  of  a  triangle,  to  construct  the 
triangle. 

m 


Given  m,  n,  and  p  the  three  sides  of  a  A. 

Required  to  construct  the  A. 

Construction.     Draw  line  AB  =  m. 

With  ^  as  a  centre  and  n  as  a  radius,  desc^ribe  an  arc ; 
with  5  as  a  centre  and  j9  as  a  radius,  describe  an  arc  inter- 
secting the  former  arc  at  C,  and  draw  lines  AC  and  BC. 

Then,  ABG  is  the  required  A. 


112  PLANE   GEOMETRY.  — BOOK  II. 

217.  Sch.  The  problem  is  impossible  when  one  of  the 
given  sides  is  equal  to,  or  greater  than,  the  sum  of  the  other 
two.  (§  61) 

Prop.  XXXVIII.     Problem. 

218.  Given  two  sides  of  a  triangle,  and  the  angle  opposite 
to  one  of  them,  to  construct  the  triangle. 

Given  m  and  n  two  sides  of  a  A,  and  A'  the  Z  opposite 
to  n. 

Required  to  construct  the  A. 

Construction.  Construct  Z  DAE  =  ZA'  (§  208),  and  on 
AE  take  AB  =  m. 

With  ^  as  a  centre  and  n  as  a  radius,  describe  an  arc. 

Case  I.    When  A'  is  acute,  and  m>  n. 

There  may  be  three  cases : 

1.   The  arc  may  intersect  AD  in  two  points. 


Let  Ci  and  Cg  be  the  points  in  which  the  arc  intersects 
AD,  and  draw  lines  BCi  and  BG2' 

Then,  either  ABCi  or  ABC2  is  the  required  A. 

Note.    This  is  called  the  ambiguous  case. 

2.  The  arc  may  be  tangent  to  AD. 
In  this  case  there  is  but  one  A. 

And  since  a  tangent  to  a  O  is  JL  to  the  radius  drawn 
to  the  point  of  contact  (§  170),  the  A  is  a  right  A. 

3.  The  arc  may  not  intersect  AD  at  all. 
In  this  case  the  problem  is  impossible. 
Case  II.    When  A^  is  acute,  and  m  —  n. 


THE   CIRCLE.  113 

In  this  case,  the  arc  intersects  AD  in  two  points,  one 
of  which  is  A. 

Then  there  is  but  one  A ;  an  isosceles  A. 

Case  m.    WJien  A'  is  acute,  and  m<n. 


m 


In  this  case,  the  arc  intersects  AD  in  two  points. 

Let  Ci  and  C2  be  the  points  in  which  the  arc  intersects 
AD,  and  draw  lines  BCi  and  BC^. 

Now  A  ABCi  cloes  not  satisfy  the  conditions  of  the  prob- 
lem, since  it  does  not  contain  the  given  Z  A'. 

Then  there  is  but  one  A ;  A  ABC2. 

Case  rV.    When  A'  is  right  or  obtuse,  and  m<n. 
In  each  of  these  cases,  the  arc   intersects  AD  in  two 
points  on  opposite  sides  of  A. 
Then  there  is  but  one  A. 

219.  Sch.  If  A'  is  right  or  obtuse,  and  m  =  n  or  m>n, 
the  problem  is  impossible ;  for  the  side  opposite  the  right 
or  obtuse  angle  in  a  triangle  must  be  the  greatest  side  of 
the  triangle.  (§  99) 

The  pupil  should  construct  the  triangle  corresponding 
to  each  case  of  §  218. 

EXERCISES. 

78.  Given  one  of  the  equal  sides  and  the  altitude  of  an  isosceles 
triangle,  to  construct  the  triangle. 

What  restriction  is  there  on  the  values  of  the  given  lines  ? 

79.  Given  two  diagonals  of  a  parallelogram  and  their  included 
angle,  to  construct  the  parallelogram.     (§  111.) 


114  PLANE   GEOMETRY.  — BOOK  II. 


Prop.  XXXIX.     Problem. 

220.   Oiven  two  sides  and  the  included  angle  of  a  parallelo- 
gram, to  construct  the  parallelogram. 


m 


Given  m  and  n  two  sides,  and  A'  the  included  Z,  of  a  O. 
(The  construction  and  proof  are  left  to  the  pupil.) 

Prop.  XL.     Problem. 
221.    To  inscnhe  a  circle  in  a  given  triangle. 


^^ ^-^      '^g 

Given  A  ^5(7. 

Required  to  inscribe  a  O  in  A  ABC. 
Construction.     Draw  lines  AD  and  BE  bisecting  A  A  and 
B,  respectively  (§  207). 

From  their  intersection  0,  draw  line  OM  A.  AB  (§  204). 
With  0  as  a  centre  and  OM  as  a  radius,  describe  a  O. 
This  O  will  be  tangent  to  AB,  BC,  and  CA. 
(The  proof  is  left  to  the  pupil ;  see  §  135.) 


Ex.  80.   To  construct  a  right  triangle,  having  given  the  hypotenuse 
and  an  acute  angle. 

(The  other  acute  Z  is  the  complement  of  the  given  Z.) 


THE  CIRCLE.  II5 

Prop.  XLI.    Problem. 
222.   To  circumscribe  a  circle  about  a  given  triangle. 

C 


Given  A  ABC. 

Required  to  circumscribe  a  O  about  A  ABC. 

Construction.     Draw  lines  DF  and  EG  ±  to  AB  and  AC, 

respectively,  at  their  middle  points  (§  205). 
Let  DF  and  EO  intersect  at  0. 

With  0  as  a  centre,  and  OA  as  a  radius,  describe  a  O. 
The  circumference  will  pass  through  A,  B,  and  G. 
(The  proof  is  left  to  the  pupil ;  see  §  137.) 

223.  Sch.  The  above  construction  serves  to  describe  a 
circumference  through  three  given  points  not  in  the  same 
straight  line,  or  to  find  the  centre  of  a  given  circumference 
or  arc. 

EXERCISES. 

81.  To  construct  a  right  triangle,  having  given  a  leg  and  the  oppo- 
site acute  angle. 

(Construct  the  complement  of  the  given  Z.) 

82.  Given  the  base  and  the  vertical  angle  of  an  isosceles  triangle, 
to  construct  the  triangle, 

(Each  of  the  equal  A  is  the  complement  of  one-half  the  vertical  Z.) 

83.  Given  the  altitude  and  one  of  the  equal  angles  of  an  isosceles 
triangle,  to  construct  the  triangle. 

(One-half  the  vertical  Z  is  the  complement  of  each  of  the  equal  A.) 

84.  To  circumscribe  a  circle  about  a  given  rectangle^ 
(Draw  the  diagonals.) 


116 


PLANE   GEOMETRY.  — BOOK  II. 


Prop.  XLII.     Problem. 

224.    To  draw  a  tangent  to  a  circle  through  a  given  point 
on  the  circumference. 

BAG 


Given  A  any  point  on  the  circumference  of  O  AD. 
Required  to  draw  through  A  a  tangent  to  O  AD. 
Construction.     Draw  radius  OA. 
Through  A  draw  line  BC  ±  OA  (§  203). 
Then,  BG  will  be  tangent  to  O  AD. 


(?) 


Prop.  XLIII.     Problem. 

225.    To  draw  a  tangent  to  a  circle  through  a  given  point 
without  the  circle. 


Given  A  any  point  without  O  BC. 

Required  to  draw  through  A  a  tangent  to  O  BG. 

Construction.  Let  0  be  the  centre  of  Q  BG,  and  draw 
line  OA. 

On  OA  as  a  diameter,  describe  a  circumference,  cutting 
the  given  circumference  at  B  and  G. 

Draw  lines  AB  and  AG. 


THE   CIRCLE. 


117 


Then,  AB  and  AC  are  tangents  to  O  BC. 

Proof.     Draw  line  OB. 

Z  ABO,  being  inscribed  in  a  semicircle,  is  a  rt.  Z. 

Therefore,  AB  is  tangent  to  O  BC. 

In  like  manner,  ^C  is  tangent  to  O  BC. 


(?) 
(?) 


Prop.  XLIV.    Problem. 

226.    Upon  a  given  straight  line,  to  descHbe  a  segment  which 
shall  contain  a  given  angle. 

a 


•■■■^ 

> 

(F 

v.    J 

r 

B 


Given  line  AB,  and  Z  A'. 

Required  to  describe  upon  AB  a  segment  such  that  every 
Z  inscribed  in  the  segment  shall  equal  Z  A'. 

Construction.     Construct  Z  BAC  =  Z  A'.  (§  208) 

Draw  line  DE  ±  to  AB  at  its  middle  point.  (§  205) 

Draw  line  AFl.  AC,  intersecting  DE  at  0. 
With  0  as  a  centre  and  OA  as  a  radius,  describe  O  AMBN. 
Then,  AMB  will  be  the  required  segment. 

Proof,     Let  AOB  be  any  Z  inscribed  in  segment  AMB. 


Then,  Z  AGB  is  measured  by  \  arc  ANB. 
But,  by  cons.,  AC  _L  OA. 
Whence,  ^C  is  tangent  to  O  AMB. 
Therefore,  Z  BAC  is  measured  by  ^  arc  ANB. 

.'.  ZAGB  =  ZBAC  =  ZA'. 


(?) 


(?) 
(§  197) 

(?) 

Hence,  every  Z  inscribed  in  segment  AMB  equals  Z  A'. 

(§  194) 


118 


PLANE   GEOMETRY.  — BOOK  II. 


EXERCISES. 

85.  Given  the  middle  point  of  a  chord  of  a  circle, 
to  construct  the  chord. 

(To  draw  through  C  a  chord  which  is  bisected 
at  C.) 


86.    To  draw  a  line  tangent  to  a  given  circle  and 
parallel  to  a  given  straight  line. 
(To  draw  a  tangent  ||  AB.) 


87.  To  draw  a  line  tangent  to  a  given  circle  and  perpendicular  to 
a  given  straight  line. 

88.  To  draw  a  straight  line  through  a  given  point 
within  a  given  acute  Z,  forming  with  the  sides  of     A 
the  angle  an  isosceles  triangle. 

89.  Given  the  base,  an  adjacent  angle,  and  the  altitude  of  a  tri- 
angle, to  construct  the  triangle. 

(Draw  a  II  to  the  base  at  a  distance  equal  to  the  altitude,) 

90.  Given  the  base,  an  adjacent  side,  and  c^-— 
the  altitude   of   a  triangle,  to  construct  the        ^,--''^S^iD 
triangle.  n\p 

Discuss  the  problem  for  the  following  cases  : 

A 


1.    n>p. 


3.    n<p. 


91.  To  construct  a  rhombus,  having  given  its  base  and  altitude. 
(Draw  a  II  to  the  base  at  a  distance  equal  to  the  altitude.) 
What  restriction  is  there  on  the  values  of  the  given  lines  ? 

92.  Given  the  altitude  and  the  sides  includ- 
ing the  vertical  angle  of  a  triangle,  to  construct 
the  triangle. 

What  restriction  is  there  on  the  values  of 
the  given  lines  ? 

Discuss  the  problem  for  the  following  cases : 
1.   m  <  w  or  >  w.  2.   m  —  n. 


THE  CIRCLE.  219 

93.  Given  the  altitude  of  a  triangle,  and  the  angles  at  the  extremi- 
ties of  the  base,  to  construct  the  triangle. 

(The  Z  between  the  altitude  and  an  adjacent  side  is  the  complement 
of  the  Z  at  the  extremity  of  the  base,  if  acute,  or  of  its  supplement,  if 
obtuse.) 

94.  To  construct  an  isosceles  triangle,  having  given  the  base  and 
the  radius  of  the  circumscribed  circle. 

What  restriction  is  there  on  the  values  of  the  given  lines  ? 

95.  To  construct  a  square,  having  given  one  of  its  diagonals. 
(§  195.) 

96.  To  construct  a  right  triangle,  having 

given  the  hypotenuse  and  the  length  of  the  d...^  <■" "^n 

perpendicular  drawn  to  it  from  the  vertex  of         i  //^...^      "\ 
the  right  angle.  i//  ^^""-^^  \ 

What  restriction  is  there  on  the  values  of  A*^  ^ ^^B 

the  given  lines  ? 

97.  To  construct  a  right  triangle,  having  given  the  hypotenuse  and 
a  leg. 

What  restriction  is  there  on  the  values  of  the  given  lines  ? 

98.  Given  the  base  of  a  triangle  and  the       ^  E, 
perpendiculars  from  its  extremities  to  the  other      ~"'*>s^,  c,-f'^ 
sides,  to  construct  the  triangle.     (§  225.) 

What  restriction  is  there  on  the  values  of 
the  given  lines  ? 

99.  To  describe  a  circle  of  given  radius  tangent  to  two  given 
intersecting  lines. 

(Draw  a  II  to  one  of  the  lines  at  a  distance  equal  to  the  radius.) 

100.  To  describe  a  circle  tangent  to  a  given  straight  line,  having 
its  centre  at  a  given  point  without  the  line. 

101.  To  construct  a  circle  having  its  centre  in  a  given  line,  and 
passing  through  two  given  points  without  the  line.     (§  163.) 

What  restriction  is  there  on  the  positions  of  the  given  points  ? 

102.  In  a  given  straight  line  to  find  a  point  equally  distant  from 
two  given  intersecting  lines.     (§  101.) 

103.  Given  a  side  and  the  diagonals  of  a  parallelogram,  to  con- 
struct the  parallelogram. 

What  restriction  is  there  on  the  values  of  the  given  lines  ? 

104.  Through  a  given  point  without  a  given  straight  line,  to 
describe  a  circle  tangent  to  the  given  line  at  a  given  point.     (§  163.) 


120  PLANE   GEOMETRY.— BOOK  II. 


105.  Through  a  given  point  within  a  circle  to 
draw  a  chord  equal  to  a  given  chord.     (§  164.) 

What  restriction  is  there  on  the  position  of  the 
given  point? 


106.  Through  a  given  point  to  describe  a  circle  of  given  radius 
tangent  to  a  given  straight  line. 

(Draw  a  ||  to  the  given  line  at  a  distance  equal  to  the  radius.) 

107.  To  describe  a  circle  of  given  radius 
tangent  to  two  given  circles. 

(To  describe  a  O  of  radius  m  tangent  to 
two  given  (D  whose  radii  are  n  and  p,  respec- 
tively.) 

What  restriction  is  there  on  the  value  of  m  ? 

108.  To  describe  a  circle  tangent  to  two  given  parallels,  and  pass- 
ing through  a  given  point. 

What  restriction  is  there  on  the  position  of  the  given  point  ? 

109.  To  describe  a  circle  of  given  radius,  tangent  to  a  given  line 
and  a  given  circle. 

(Draw  a  II  to  the  given  line  at  a  distance  equal  to  the  given  radius.) 

110.  To  construct  a  parallelogram,  having  given  a  side,  an  angle, 
and  the  diagonal  drawn  from  the  vertex  of  the  angle. 

111.  In  a  given  triangle  to  inscribe  a  rhombus,  having  one  of  its 
angles  coincident  with  an  angle  of  the  triangle. 

(Bisect  the  Z  which  is  common  to  the  A  and  the  rhombus.) 

112.  To  describe  a  circle  touching  two  given  intersecting  lines, 
one  of  them  at  a  given  point.     (§  169.) 

113.  In  a  given  sector  to  inscribe  a  circle. 
(The  problem  is  the  same  as  inscribing  a  O  in 

A  O'CD.) 


114.  In  a  given  right  triangle  to  inscribe  a  square,  having  one 
of  its  angles  coincident  with  the  right  angle  of  the  triangle. 

115.  Through  a  vertex  of  a  triangle  to  draw  a  straight  line  equally 
distant  from  the  other  vertices. 


THE  CIRCLE. 


121 


116.  Given  the  base,  the  altitude,  and  the  vertical  angle  of  a  tri- 
angle, to  construct  the  triangle.     (§  226. ) 

(Construct  on  the  given  base  as  a  chord  a  segment  which  shall 
contain  the  given  Z. ) 

117.  Given  the  base  of   a  triangle,  its  vertical  angle,  and  the 
median  drawn  to  the  base,  to  construct  the  triangle. 


118.   To  construct  a  triangle,  having  given  the 
middle  points  of  its  sides. 


119.  Given  two  sides  of  a  triangle,  and  the 
median  drawn  to  the  third  side,  to  construct 
the  triangle. 

(Construct  AABD  with  its  sides  equal  to  g 
m,  w,  and  2p,  respectively.) 

What  restriction  is  there  on  the  values  of 
the  given  lines  ? 

120.  Given  the  base,  the  altitude,  and  the  radius  of  the  circum- 
scribed circle  of  a  triangle,  to  construct  the  triangle. 

(The  centre  of  the  circumscribed  O  lies  at  a  distance  from  each 
vertex  equal  to  the  radius  of  the~0.) 

121.  To  draw  common  tangents  to  two  given  circles  which  do  not 
intersect. 


(To  draw  exterior  common  tangents,  describe  O  AA'  with  its  radius 
equal  to  the  difference  of  the  radii  of  the  given  (D. 

To  draw  interior  common  tangents,  describe  O  AA'  with  its  radius 
equal  to  the  sum  of  the  radii  of  the  given  (D.) 


Note.     For  additional  exercises  on  Book  II.,  see  p.  224. 


Book  III. 


THEORY   OP   PROPORTION. -SIMILAR 
POLYGONS. 

DEFINITIONS. 

227.  A  Proportion  is  a  statement  that  two  ratios  are 
equal. 

228.  The  statement  that  the  ratio  of  a  to  5  is  equal  to  the 
ratio  of  c  to  d,  may  be  written  in  either  of  the  forms 

a  :  0  =  c  :  a,  or  -  =  — 
b      d 

229.  The  first  and  fourth  terms  of  a  proportion  are  called 
the  extremes,  and  the  second  and  third  terms  the  means. 

The  first  and  third  terms  are  called  the  antecedents,  and 
the  second  and  fourth  terms  the  consequents. 

Thus,  in  the  proportion  a'.h  =  c:d,  a  and  d  are  the 
extremes,  h  and  c  the  means,  a  and  c  the  antecedents,  and 
h  and  d  the  consequents. 

230.  If  the  means  of  a  proportion  are  equal,  either  mean 
is  called  a  mean  proportional  between  the  first  and  last  terms, 
and  the  last  term  is  called  a  third  proportional  to  the  first 
and  second  terms. 

Thus,  in  the  proportion  a:h  —  h:c,h  is  a  mean  propor- 
tional between  a  and  c,  and  c  a  third  proportional  to  a  and  h. 

231.  A  fourth  proportional  to  three  quantities  is  the 
fourth  term  of  a  proportion,  whose  first  three  terms  are  the 
three  quantities  taken  in  their  order. 

Thus,  in  the  proportion  a:h  =  c:  d,  d  is  a  fourth  propor- 
tional to  a,  b,  and  c. 

122 


THEORY  OF  PROPORTION.  123 

Prop.  I.    Theorem. 

232.  In  any  proportion,  the  product  of  the  extremes  is  equal 
to  the  product  of  the  means. 

Given  the  proportion  a:b  =  c:d. 

To  Prove  ad  =  be. 

Proof.     By  §228,  -  =  -^. 

b     d 

Multiplying  both  members  of  this  equation  by  bd, 

ad  =  be. 

233.  Cor.  The  mean  proportional  between  two  quantities 
is  equal  to  the  square  root  of  their  product. 

Given  the  proportion    a:b  =  b:c.  (1) 

To  Prove  b  =  Vac: 

Proof.     From  (1),  b'  =  ac.  (§  232) 

.-.  b  =^/ac. 

Prop.  II.     Theorem. 

234.  (Converse  of  Prop.  I.)  If  the  product  of  two  quan- 
tities is  equal  to  the  product  of  two  others,  one  pair  may  be 
made  the  extremes,  and  the  other  pair  the  means,  of  a 
proportion. 

Given  ad  —  be.  (1) 

To  Prove  a:b  =  c.d. 

Proof.     Dividing  both  members  of  (1)  by  bd, 

ad  _bc 

bd~bd 


Or, 

b~~d 

Then  by  §  228, 

a:b  =  c:d. 

In  like  manner, 

a:c  =  b  :  d. 

b:a  =  d:  c,  etc. 

124  PLANE   GEOMETRY.— BOOK  III. 


Prop.  III.     Theorem. 

235.   In  any  proportion,  the  terms  are  in  proportion  by 
Alternation  ;   that  is,  the  first  term  is  to  the  third  as  the 


second  term 

is  to  the  fourth. 

Given  the 

proportion 

a:b: 

=  c: 

d. 

To  Prove 

a:  c: 

=  b: 

d. 

Proof,    rrom  (1), 

ad: 

=  bc 

'. 

••• 

a:  c: 

=  b: 

d. 

Prop. 

IV. 

Theorem. 

(1) 

(§  232) 
(§  234) 


236.  In  any  proportion,  the  terms  are  in  proportion  by 
Inversion;  that  is,  the  second  term  is  to  the  first  as  the 
fourth  term  is  to  the  third. 

Given  the  proportion    a:b  =  c:d.  (1) 
To  Prove                       b:a  =  d:  c. 

Proof.     From  (1),            ad  =  be.  (?) 

.-.  b:a  =  d:c.  (?) 

Prop.  V.     Theorem. 

237.  In  any  proportion,  the  terms  are  in  proportion  by 
Composition  ;  that  is,  the  sum  of  the  first  two  terms  is  to 
the  first  term  as  the  sum  of  the  last  two  terms  is  to  the  third 
term. 

Given  the  proportion    a:b  —  c:d.  (1) 

To  Prove  a-\-b:a  =  c-[-d:c. 

Proof.     From  (1),  ad  =  be.  (?) 

Adding  both  members  of  the  equation  to  ac, 

ac-\-  ad  =  ac  +  be. 
Factoring,  a(c  ■i-d)  =  c{a  +  b). 

.-.  a  +  b'.a  =  c-\-d'.c.  (§  234) 

In  like  manner,      a -\-b:b —  c-\- d:d. 


THEORY  OF  PROPORTION.  125 

Prop.  VI.     Theorem. 

238.  In  any  proportion,  the  terms  are  in  proportion  by 
Division  ;  that  is,  the  difference  of  the  first  two  terms  is  to 
the  first  term  as  the  difference  of  the  last  two  terms  is  to  the 
third  term. 

Given  the  proportion    a :  6  =  c :  c?,  (1) 

in  which  a>h  and  c>d. 
To  Prove  a  —  h:a=c  —  d:c. 

Proof.     From  (1),  ad  =  he.  (?) 

Subtracting  both  members  of  the  equation  from  acj 

ac  —  ad  =  ac  —  be. 
Factoring,  a(c  —  d)  =  c(a  —  b). 

.:  a  —  b  :  a  =  c  ~  d :  c.  (?) 

In  like  manner,      a  —  b:  b  =  c  —  did. 

Prop.  VII.     Theorem. 

239.  In  any  proportion,  the  terms  are  in  proportion  by 
Composition  and  Division;  that  is,  the  sum  of  the  first 
two  terms  is  to  their  difference  as  the  sum  of  the  last  two 
terms  is  to  their  difference. 

Given  the  proportion     a:b  =  c:d,  (1) 

in  which  a>b  and  c>d. 
To  Prove  a-\-b:a—b  =  c  +  d:c  —  d. 

Proof.     From(l),        a±b^c_±d^  ^g  237) 

a  c 

and  ^L^^^Jzl.  (§238) 

a  c 

Dividing  the  first  equation  by  the  second, 

a-^b  _c  +_d 

a  —  b     c  —  d 

.'.  a  +  b:a  —  b  =  c-\-d:c  —  d. 


126  PLANE   GEOMETRY. -BOOK  III. 

Prop.  VIII.     Theorem. 

240.  In  a  series  of  equal  ratios,  the  sum  of  all  the  antece- 
dents is  to  the  sum  of  all  the  consequents  as  any  antecedent 
is  to  its  consequent. 

Given  a:b  =  c:d  =  e:f  (1) 

To  Prove  a-\- c-^e-.b +  d  +/=  a :  b. 

Proof.     We  have  ba  =  ab. 

Also,  from  (1),  be  =  ad, 

and  be  =  af  (?) 

Adding,  ba  +  be  +  be=  ab  +  ad  -\-  af. 

.'.  b{a^c  +  e)  =  a{b-\-d+f). 
.'.  a-{-c-\-e:b-j-d  -f /=  a :  b.  (?) 

Prop.  IX.     Theorem. 

241.  In  any  proportion,  like  powers  or  like  roots  of  the 
terms  are  in  proportion. 

Given  the  proportion     a:b  =  c:d.  (1) 

To  Prove  a" :  6"  =  c"  :  d*". 

Proof.     From(l),  ^  =  | 

Raising  both  members  to  the  nth  power, 

b^  ~~  d^ 
.'.  a^  :  5»  =  c«  :  6?**. 
In  like  manner,      Va :  V6  =  ->/c :  -^d. 

Note.  The  ratio  of  two  magnitudes  of  the  same  kind  is  equal 
to  the  ratio  of  their  numerical  measures  when  referred  to  a  common 
unit  (§  183)  ;  hence,  in  any  proportion  involving  the  ratio  of  two 
magnitudes  of  the  same  kind,  we  may  regard  the  ratio  of  the  mag- 
nitudes as  replaced  by  the  ratio  of  their  numerical  measures  when 
referred  to  a  common  unit. 

Thus,  let  AB,  CD,  EF,  and  GH  be  four  lines  such  that 
AB:CD  =  EF:  GH. 

Then,  ABx  GH=CDx  EF.  (§  232) 


THEORY  OF  PROPORTION.  127 

This  means  that  the  product  of  the  numerical  measures  of  AB  and 
GH  is  equal  to  the  product  of  the  numerical  measures  of  CD  and  EF. 

An  interpretation  of  this  nature  must  be  given  to  all  applications 
of  §§  232,  233  and  241. 

EXERCISES. 

1.  Find  a  fourth  proportional  to  65,  80,  and  91. 

2.  Find  a  mean  proportional  between  28  and  63. 

3.  Find  a  third  proportional  to  |  and  f . 

4.  What  is  the  second  term  of  a  proportion  whose  first,  third,  and 
fourth  terms  are  45,  160,  and  224,  respectively  ? 

PROPORTIONAL  LINES. 
Prop.  X.     Theorem. 

242.   If  a  series  of  parallels,  cutting  two  straight  lines, 
intercept  equal  distances  on  one  of  these  lines,  they  also  inter- 
cept equal  distances  on  the  other. 
A 
c/ \C' 


\F" 

is' 

Given  lines  AB  and  A'B'  cut  by  lis  CC,  DD',  EE\  and 
FF^  at  points  G,  D,  E,  F,  and  O,  D',  E',  F',  respectively, 
so  that  CD  =  DE  =  EF. 

To  Prove  OD'  =  D'E'=  E'F'. 

Proof.     In  trapezoid  CC'E'E,  by  hyp.,  DD'  is  II  to  the 
bases,  and  bisects  CE ;  it  therefore  bisects  C'E'.         (§  133) 
.-.  C'D'  =  D'E'.     .  (1) 

In  like  manner,  in  trapezoid  DD'F'F,  EE'  is  II  to  the 
bases,  and  bisects  DF. 

.'.  D'E'  =  E'F'.  (2) 

From  (1)  and  (2),  CD'  =  D'E'  =  E'F'.  (?) 


128  PLANE   GEOMETRY.  — BOOK  III. 

243.  Def.  Two  straight  lines  are  said  to  be  divided  pro- 
portionally when  their  corresponding  segments  are  in  the 
same  ratio  as  the  lines  themselves.         ^         ^  ^  ^ 

Thus,  lines  AB  and  CD  are  divided   ! ^ 1 

proportionally  if  C       F  D 

AE^BE^AB 
CF      DF     CD 


Prop.  XI.     Theorem. 

244.   A  parallel  to  one  side  of  a  triangle  divides  the  other 
two  sides  proportionally. 

Case  I.      Wlien  the  segments  of  each  side  are  commensvr 
rdble. 

A 

fA\ 


Given,  in  A  ABC,  segments  AD  and  BD  of  side  AB 
commensurable,  and  line  DE  II  BC,  meeting  AC  at  E. 

To  Prove  ^  =  ^. 

BD      CE 

Proof.     Let  AF  be  a  common  measure  of  AD  and  BD ; 

and  let  it  be  contained  4  times  in  AD,  and  3  times  in  BD. 

...^  =  i  (1) 

BD     3  ^  ^ 

Drawing  lis  to  BC  through  the  several  points  of  division 
of  AB,  AE  will  be  divided  into  4  parts,  and  CE  into  3 
parts,  all  of  which,  parts  are  equal.  (§  242) 

.-.  ^  =  i  (2) 

From  (1)  and  (2),         |§  =  g.  (?) 


THEORY  OF  PROPORTION.  129 

Case  n.      When  the  segments  of  each  side  are  incommenr 
surahle. 

A 


Given,  in  A  ABC,  segments  AD  and  BD  of  side  AB 
incommensurable,  and  line  DE  II  BC,  meeting  AG  at  E. 

To  Prove  bD=CE' 

Proof.  Let  AD  be  divided  into  any  number  of  equal 
parts,  and  let  one  of  these  parts  be  applied  to  BD  as  a 
unit  of  measure. 

Since  AD  and  BD  are  incommensurable,  a  certain  num- 
ber of  the  equal  parts  will  extend  from  D  to  B',  leaving  a 
remainder  BB'  <  one  of  the  equal  parts. 

Draw  line  B'C  II  BO,  meeting  ^C  at  C. 

Then,  since  AD  and  B'D  are  commensurable, 

|g  =  g.  (§244,  Case  I.) 

Now  let  the  number  of  subdivisions  of  AD  be  indefinitely 
increased. 

Then  the  unit  of  measure  will  be  indefinitely  diminished, 
and  the  remainder  BB'  will  approach  the  limit  0. 

Then,  — — -  will  approach  the  limit  —-, 

B'D  ^^  BD 

and  will  approach  the  limit  — — • 

C'E  ^^  CE 

By  the  Theorem  of  Limits,  these  limits  are  equal.         (?) 

AD^AE 

' '  BD      CE 


130  PLANE   GEOMETRY.— BOOK  III. 

245.  Cor.  I.     We  may  write  the  result  of  §  244, 

AD:BD  =  AE:CE.  (1) 

.-.  AD  +  BD'.AD  =  AE+  CE :  AE.  (§  237) 

.-.  AB:AD=^AC'.AE.  (2) 

In  like  manner,    AB:BD  =  AC:  CE.  (3) 

246.  Cor.  n.     From  (2),  §245,. 

AB.AC  =  AD:AE, 

and  from  (3),  AB :  AC  =  BD  :  CE.  (§  235) 

mi,       u     A      i         AB     AD     BD  x.x 

Then,  by  Ax.  1,       _  =  _  =  _.  (4) 

247.  Sch.     The    proportions    (1),   (2),   (3),   and   (4),   of 
§  §  245  and  246,  are  all  included  in  the  general  statement, 

A  parallel  to  one  side  of  a  triangle  divides  the  other  two 
sides  proportionally. 

Prop.  XII.     Theorem. 

248.  (Converse  of  Prop.  XI.)     A  line  which  divides  two 
sides  of  a  triangle  proportionally  is  parallel  to  the  third  side. 

A 


B  C 

Given,  in  A  ABC,  line  DE  meeting  AB  and  AC  at  D  and 

E  respectively,  so  that 

AB^AC 

AD     AE 

To  Prove  DE  II  BG. 

Proof.     If  DE  is  not  II  BC,  draw  line  DF  II  BC,  meeting 

AC  at  F. 

,.,4^  =  A£.  (§247) 

AD     AF  ^\     ^ 


(?) 


THEORY  OF  PROPORTION.  131 

But  by  hyp.,  4^  =  A^, 

^     ^^'  AD     AE 

AC^AG 

' '  AE     AF 
.-.  AE  =  AF. 

Then,  DF  coincides  with  DE,  and  DE  II  BC.  (Ax.  3) 

Prop.  XIII.     Theorem. 

249.   In  any  ttiangle,  the  bisector  of  an  angle  divides  the 
opposite  side  into  segments  proportional  to  the  adjacent  sides. 
E 


B  D 

Given  line  AD  bisecting  Z  J.  of  A  ABC,  meeting  BC  at  D. 

Proof.     Draw  line  BE  II  AD,  meeting  CA  produced  at  E. 
Then,  since  lis  AD  and  BE  are  cut  by  AB, 

ZABE  =  ZBAD.  (?) 

And  since  lis  AD  and  BE  are  cut  by  CE, 

Z  AEB  =  Z  CAD.  (?) 

But  by  hyp.,  ZBAD  =  Z  CAD. 

.-.  ZABE  =  ZAEB.  (?) 

.-.  AB  =  AE.  (?) 

Now  since  AD  is  II  to  side  BE  of  A  BCE, 

M  =  4R.  (§247) 

DC     AC  ^         ^ 

Putting  for  AE  its  equal  AB,  we  have 

DB^AB 

DC     AG 


132  PLANE   GEOMETRY.  — BOOK  III. 

250.  Def.  The  segments  of  a  line  by  a  point  are  the  dis- 
tances from  the  point  to  the  extremities  of  the  line,  whether 
the  point  is  in  the  line  itself,  or  in  the  line  produced. 

Prop.  XIV.     Theorem. 

251.  In  any  triangle  the  bisector  of  an  exterior  angle  divides 
the  opposite  side  externally  into  segments  proportional  to  the 
adjacerit  sides. 

Note.    The  theorem  does  not  hold  for  the  exterior  angle  at  the 
vertex  of  an  isosceles  triangle. 
E 


D  B 

Given  line  AD  bisecting  ext.  Z  BAE  of  A  ABC,  meeting 
GB  produced  at  D. 

To  Prove  ^  =  ^. 

DC     AC 

(Draw  BF  II  AD ;  then  Z  ABF=Z  AFB,  and  AF=AB; 

BF  is  II  to  side  AD  of  A  ACD.) 

SIMILAR  POLYGONS. 

252.  Def.     Two  polygons  are  said  to  be  similar  if  they 
are  mutually  equiangular  (§  122),  and  have  their  homolo- 
gous sides  (§  123)  proportional. 
B 


E  D  E'  D 

Thus,  polygons  ABCDE  and  A'B'C'D'F'  are  similar  if 

ZA=ZA',  ZB  =  ZB',etG., 
,  AB       BC       CD      . 

""^'  A^'^WC'-^Cr^'"''' 


SIMILAR  POLYGONS.  133 

253.  Sch.     The  following  are  given  for  reference : 

1.  In  similar  polygons,  the  homologous  angles  are  equal. 

2.  In  similar  polygons,  the  homologous  sides  are  propor- 
tioned. 

Prop.  XV.     Theorem. 

254.  Two  triangles  are  similar  when  they  are  mutually 
equiangular. 

A 


B  ■  U 

Given,  in  A  ABC  and  A'B'O, 

ZA  =  ZA',  ZB  =  ZB',3in([ZC=ZC'. 
To  Prove  A  ABC  and  A'B'O  similar. 
Proof.     Place  AA'B'C  in  the  position  ADE;  Z  A'  coin- 
ciding with  its  equal  Z  A,  vertices  B'  and  C  falling  at  D 
and  E,  respectively,  and  side  B'C  at  DE. 

Since,  by  hyp.,  Z  ADE  =  Z  B,  DE  II  BC.  (?) 

.-.—=—.  (§247) 

AD     AE  ^^        ^ 

That  IS,  A^'  =  ^C''  (^> 

In  like  manner,  by  placing  A  A' B'C  so  that  Z  B'  shall 

coincide  with  its  equal  Z  B,  vertices  A'  and  C  falling  on 

sides  AB  and  BC,  respectively,  we  may  prove 

AB  ^  BC 

A'B'     B'C' 


(2) 


From  (1)  and  (2),         -^,  =  -^,  =  j^,-  (?) 

Then,  A  ABC  and  A' B'C  have  their  homologous  sides 
proportional,  and  are  similar.  (§  252) 


134 


PLANE   GEOMETRY.  — BOOK  III. 


255.  Cor.  I.     Tico  triangles  are  similar  when  two  angles  of 
one  are  equal  respectively  to  two  angles  of  the  other. 

For  their  remaining  A  are  equal  each  to  each.  (§  86) 

256.  Cor.  II.      Two  nght  triangles  are  similar  ivhen  an 
acute  angle  of  one  is  equal  to  an  acute  aiigle  of  the  other. 

257.  Cor.  III.     If  a  line  he  drawn  between  two  sides  of  a 
triangle  iiavallel  to  the  third  side,  the  tri- 
angle  formed   is   similar   to    the    given 
triangle. 

Given  line  DE  II  to  side  BG  of 
A  ABC,  meeting  AB  and  AC  at  D 
and  E,  respectively. 

To  Prove  A  ADE  similar  to  A  ABC. 

(The  A  are  mutually  equiangular.) 

258.  Sch.     I7i  similar  triangles,  the  homologous  sides  lie 
opposite  the  equal  angles. 

Prop.  XVI.     Theorem. 

259.  Two   triangles  are    siynilar  when   their   homologous 
sides  are  proportional. 

A 


B  C 

Given,  in  A  ABC  and  A'B'C, 
AB       AC 


BC 
B'O' 


A'B'     A'C 
To  Prove  A  ABC  and  A'B'C  similar. 
Proof.     On  AB  and  AC,  take  AD  =  A'B'  and  AE  =  A'C 
Praw  line  DJEJ ;  then,  from  the  given  proportion, 
AB^AC 
AD     AE 


SIMILAR  POLYGONS.  I35 

.-.  DE  II  BO.  (§  248) 

Then,  A  ADE  and  ABC  are  similar.  (§  257) 

.    AB     BG   ^^    AB      BO  ,.  ^kq  ox 

''AD='DE^'''A^'^nE'  ^^^'^'^^ 

1.  .V.    T.  AB       BC 

But  by  hyp.,  'aB^^B^^' 

.-.  DE  =  B'a. 
.'.  A  ADE  =  A  A'B'O.  (§  69) 

But,  A  ADE  has  been  proved  similar  to  A  ABO. 
Hence,  AA'B'O  is  similar  to  A  ^50. 

260.  Sch.  To  prove  that  two  polygons  in  general  are 
similar,  it  must  be  shown  that  they  are  mutually  equiangu- 
lar, and  have  their  homologous  sides  proportional  (§  252) ; 
but  in  the  case  of  two  triangles,  each  of  these  conditions 
involves  the  other  (§§  254,  259),  so  that  it  is  only  neces- 
sary to  show  that  one  of  the  tests  of  similarity  is  satisfied. 

Prop.  XVII.     Theorem. 

261.  Two  triangles  are  similar  when  they  have  an  angle  of 
one  equal  to  an  angle  of  the  other,  and  the  sides  including  these 
angles  proportional. 


B  C 

Given,  in  A  ABO  and  A'B'  O', 

To  Prove  A  ABO  and  A'B'O'  similar. 
(Place  A  A'B'O'  in  the  position  ADE ;  by  §  248,  DE  II  BO', 
the  theorem  follows  by  §  257.) 


136 


PLANE   GEOMETRY.  — BOOK  III. 


Prop.  XVIII.     Theorem. 

262.    Two  triangles  are  similar  when  their  sides  are  paral- 
lel each  to  each,  or  perpendicular  each  to  each. 


Fig.  1. 


Fig.  S. 


Given  sides  AB,  AC,  and  BC,  of  A  ABC,  II  respectively 
to  sides  A'B',  A'C,  and  B'C  of  AA'B'C  in  Pig.  2,  and  ± 
respectively  to  sides  A'B',  A'C,  and  B'C  of  AA'B'C  in 
Fig.  3. 

To  Prove  A  ABC  and  A' B'C  similar. 

Proof.  Since  the  sides  of  A  A  and  A'  are  II  each  to  each, 
or  _L  each  to  each,  A  A  and  A'  are  either  equal  or  supple- 
mentary. "  (§§  81,  82,  83) 

In  like  manner,  A  B  and  B',  and  A  C  and  C,  are  either 
equal  or  supplementary. 

We  may  then  make  the  following  hypotheses  with  regard 
to  the  A  of  the  A : 

1.  ^  +  ^'  =  2rt.  A,  B  +  B'  =  2Tt.  A,  C+C'  =  2rt  A. 

2.  A-\-A'  =  2  rt.  A,  B  +  B'  =  2  rt.  A,  C=  C. 

3.  A-\-A'  =  2vt  A,  B  =  B',  (7+C"  =  2rt.A 

4.  A  =  A',  B  +  B'  =  2Tt  A,   C+C  =  2Tt.A. 

5.  A  =  A',  B=B',    whence    C=C.   (§86) 

The  first  four  hypotheses  are  impossible;  for,  in  either 
case,  the  sum  of  the  six  A  of  the  two  A  would  be  >  4  rt.  A. 

(§  84) 
We  can  then  have  only  A  =  A',B  =  B',  and  C  =  C. 
Therefore,  A  ABC  and  A'B'C  are  similar.  (§  254) 


SIMILAR  POLYGONS. 


137 


263.  Sch.     1.  In  similar  triangles  whose  sides  are  parallel 
each  to  each,  the  parallel  sides  are  homologous. 

2.   In  similar  tnangles  whose  sides  are  perpendicular  each 
to  each,  the  perpendicular  sides  are  homologous. 

Prop.  XIX.    Theorem. 

264.  The  homologous  altitudes  of  two  similar  triangles  are 
in  the  same  ratio  as  any  two  homologous  sides. 

A 


B       ^D  C  B'     ly  c 

Given  AD  and  A'D'  homologous   altitudes  of   similar 
A  ABC  and  A'B'C. 

AD  ^  AB  ^AO  ^  BO 
B'C' 


To  Prove 


A'D'     A'B'     A'C 
(Rt.  AABD  and  A'B'D'  are  similar  by  §  256.) 

265.  Sch.  In  two  similar  triangles,  any  two  homologous 
lines  are  in  the  same  ratio  as  any  two  homologous  sides. 

EXERCISES. 

5.  The  sides  of  a  triangle  are  AB  =  S,  BC=6^  and  CA=7 ;  find 
the  segments  into  which  each  side  is  divided  by  the  bisector  of  the 
opposite  angle. 

6.  The  sides  of  a  triangle  are  AB  =  5,  BC—7,  and  CA=8;  find 
the  segments  into  which  each  side  is  divided  by  the  bisector  of  the 
exterior  angle  at  the  opposite  vertex. 

7.  The  sides  of  a  triangle  are  5,  7,  and  9.  The  shortest  side  of 
a  similar  triangle  is  14.     What  are  the  other  two  sides  ? 

8.  Two  isosceles  triangles  are  similar  when  their  vertical  angles 
are  equal.     (§  255.) 

9.  The  base  and  altitude  of  a  triangle  are  5  ft.  10  in.  and  3  ft. 

6  in.,  respectively.     If  the  homologous  base  of  a  similar  triangle  is 

7  ft.  6  in. ,  find  its  homologous  altitude. 


138  PLANE   GEOMETRY.  — BOOK  III. 


Prop.  XX.     Theorem. 

266.  Two  polygons  are  similar  when  they  are  composed  of 
the  same  number  of  triangles,  similar  each  to  each,  and 
similarly  placed. 


Given,   in  polygons   AC  and  A'C,  A  ABE   similar   to 
AA'B'E',  ABCE  to  AB'C'E',  and  A  CDE  to  A  C'D'E'. 
To  Prove  polygons  AC  and  A'C  similar. 
Proof.     Since  A  ABE  and  A'B'E'  are  similar, 

ZA  =  ZA'.  (?) 

Also,  *  Z  ABE  =  Z  A'B'E'. 

And  since  A  BCE  and  B'C'E'  are  similar, 
ZEBC=ZE'B'C'. 
. :  Z  ABE  +  Z  EBC  =  Z  A'B'E'  +  Z  E'B'O. 
Or,  ZABC  =  ZA'B'C'. 

In  like  manner,       Z  BCD  =  Z  B'C'D',  etc. 
Then,  AC  and  A'C  are  mutually  equiangular. 
Again,  since  A  ABE  is  similar  to  A  A'B'E',  and  A  BCE 
to  AB'C'E', 

^^       ^^    and    ^^  =  ^.  (?) 

TDI  TTll  -Dim  V     / 


A'B'     B'E'  B'E'     B'C 

AB  ^  BC 
A'B'     B'C' 

In  like  manner,  ^  =  ^^  =  -^,  etc. 

A'B'      B'C      CD'' 


(?) 


Then,  AC  and  A'C  have  their  homologous  sides  propor- 
tional. 

Therefore,  AC  and  A'C  are  similar.  (§  252) 


SIMILAR  POLYGONS.  I39 

Prop.  XXI.     Theorem. 

267.  (Converse  of  Prop.  XX.)  Two  similar  polygons 
may  be  decomposed  into  the  same  number  of  triangles,  similar 
each  to  each,  and  sim.i-^^  ly  placed. 


Given  E  and  E^  homologous  vertices  of  similar  polygons 
AC  and  A'C,  and  lines  EB,  EC,  E'B',  and  E'C. 

To  Prove  A  ABE  similar  to  A  A'B'E',  A  BGE  to 
A  B'C'E',  and  A  CDE  to  A  C'D'E'. 

Proof.     Since  polygons  AG  and  A'C  are  similar, 

Z^  =  Z^'  and  4^  =  ^.  (?) 

A'E'      A'B'  ^  ' 

Then,  A  ABE  and  A!B'E'  are  similar.  (§  261) 

Again,  since  the  polygons  are  similar, 

Z.ABC=/.AB^a. 

And  since  A  ABE  and  A'B'E'  are  similar, 

Z  ABE  =  Z  A'B'E'. 

.'.  Z  ABC  -  Z  ABE  =  Z  A'B'C  -  Z  A'B'E'. 

Or,  ZEBC=ZE'B'C'. 

AB       BC 


Also,  since  the  polygons  are  similar. 


And  since  A  ABE  and  A'B'E'  are  similar. 


A'B'     B'C 

AB       BE 


A'B'     B'E 
BC       BE 


B'C     B'E' 


(?) 


Then,  since  Z  EBC  =  Z  E'B'C,  and  J^  =  ^,  A  BCE 

and  B'C'E'  are  similar.  (?) 

In  like  manner,  we  may  prove  A  CDE  and  C'D'E'  similar. 


140 


PLANE   GEOMETRY— BOOK   III. 


Prop.  XXII.     Theorem. 

268,    Tlie  perimeters  of  tivo  similar  polygons  are  in  the 
same  ratio  as  any  two  homologous  sides. 


Given  AB  and  A'B',  EC  and  S'C,  CD  and  C'D\  etc., 
homologous  sides  of  similar  polygons  J.(7and  AC. 
To  Prove 

AB+BC+CD  +  Qta.         AB       BO       CD 


A'B'  +  B'C'-\-G'D'-{-etG.     A'B'      B'C      CD' 
(Apply  §  240  to  the  equal  ratios  of  §  252.) 


-,  etc. 


Prop.  XXIII.     Theorem. 

269.   If  a  perpendicular  be  drawn  from  the  vertex  of  the 
right  angle  to  the  hypotenuse  of  a  right  triangle, 

I.    The  triangles  formed  are  similar  to  the  whole  triangle, 
and  to  each  other. 

11.    The  perpendicular  is  a  mean  proportional  between  the 
segments  of  the  hypotenuse. 

III.   Either  leg  is  a  mean  proportional  between  the  whole 
hypotenuse  and  the  adjacent  segment. 


A  D  B 

^iven  line  GD  J.  hypotenuse  AB  of  rt,  A  ABC. 


SIMILAR  POLYGONS.  141 

I.  To  Prove  A  ACD  and  BCD  similar  to  A  ABC,  and  to 
each  other. 

Proof.   In  rt.  A  ACD  and  ABC, 

/.A  =  ZA. 
Then,  A  ACD  is  similar  to  A  ABC.  (§  256) 

In  like  manner,  A  BCD  is  similar  to  A  ABC. 
Then,  A  ACD  and  BCD  are  similar  to  each  other,  for 
each  is  similar  to  A  ABC. 

II.  To  Prove       AD:  CD  =  CD:  BD. 
Proof.     Since  A  ACD  and  BCD  are  similar, 

Z  ACD  =  ZB3indZA  =  Z  BCD.      (§  253,  1) 

In  A  ACD  and  BCD,  AD  and  CD  are  homologous  sides, 

for  they  lie  opposite  the  equal  A  ACD  and  B,  respectively ; 

also,  CD  and  BD  are  homologous  sides,  for  they  lie  opposite 

the  equal  A  A  and  BCD,  respectively.  (§  258) 

/.  AD:CD=CD:BD.  (?) 

III.  To  Prove        AB :  AC  ==  AC :  AD. 
Proof.     Since  A  ABC  and  ACD  are  similar, 

Z  ACB  =  Z  ADC  3i,nd  ZB  =  Z  ACD.  (?) 

In  A  ABC  and  ACD,  AB  and  AC  are  homologous  sides, 
for  they  lie  opposite  the  equal  A  ACB  and  ADC,  respec- 
tively; also,  AC  and  AD  are  homologous  sides,  for  they  lie 
opposite  the  equal  A  B  and  ACD,  respectively.  .    (?) 

.-.  AB:AC=AC:AD.  (?) 

In  like  manner,       AB:BC  =  BC:  BD. 

270.  Cor.  I.  Since  an  angle  inscribed  in  a  semicircle 
is  a  right  angle  (§  195),  it  follows  that : 

If  a  perpendicular  be  drawn  from  any 
j)oint  in  the  circumference  of  a  circle  to 
a  diameter, 

1.  The  perpendicular  is  a  mean  proportional  between  the 
segments  of  the  diameter. 


142  PLANE   GEOMETRY.  — BOOK  III. 

2.  The  chord  joining  the  point  to  either  extremity  of  the 
diameter  is  a  mean  proportional  between  the  whole  diameter 
and  the  adjacent  segment. 

271.  Cor.  II.   The  three  proportions  of  §  269  give 

Cff=AD  xBD, 
AO^=ABxAD, 
and  W=ABxBD.  (?) 

Hence,  if  a  perpendicular  be  drawn  from  the  vertex  of  the 
right  angle  to  the  hypotenuse  of  a  right  triangle, 

1.  The  square  of  the  perpendicular  is  equal  to  the  product 
of  the  segments  of  the  hypoteiiuse. 

2.  The  square  of  either  leg  is  equal  to  the  product  of  the 
whole  hypotenuse  and  the  adjacent  segment. 

As  stated  in  Note,  p.  126,  these  equations  mean  that  the 
square  of  the  numerical  measure  of  CD  is  (equal  to  the 
product  of  the  numerical  measures  of  AD  and  BD,  etc. 

Prop.  XXIV.     Theorem. 

272.  In  any  right  triangle,  the  square  of  the  hypotenuse 
is  equal  to  the  sum  of  the  squares  of  the  legs. 


A  D  B 

Given  AB  the  hypotenuse  of  rt.  A  ABO. 
To  Prove  AB'  =  AC'  +  BC'. 

Proof.   Draw  line  CD  ±  AB. 
Then,  AC'  =  ABxAD, 

and  BC^  =  AB  x  BD.  (§  271,  2) 

Adding,      AC^  +  BC"  =  AB  x  (AD  +  BD)  =  AB  x  AB. 

.'.  ab^  =  ac^  +  bg\ 


SIMILAR  POLYGONS.  143 

273.  Cor.  I.     It  follows  from  §  272  that 

AC'  =  AB'-BC\  and  BC'  =  AB'-Aa'. 

That  is,  in  any  right  tnangle,  the  square  of  either  leg  is 
equal  to  the  sq^iare  of  the  hypotenuse,  minus  the  square  of 
the  other  leg. 

274.  Cor.  II.     If  ^(7  is  a  diagonal  of  square  ABCD, 
ACP  =  AB'  +  BC^  =  AB'  +  AB'.     (§272) 

.-.  AC'  =  2AB'. 
Dividing  both  members  by  AB^, 

^  =  2,or^=V2. 

Hence,  the  diagonal  of  a  square  is  incommensurable  with 
Us  side  (§  181). 

EXERCISES. 

10.  The  perimeters  of  two  similar  polygons  are  119  and  68  ;  if  a 
side  of  the  first  is  21,  what  is  the  homologous  side  of  the  second? 

11.  What  is  the  length  of  the  tangent  to  a  circle  whose  diameter 
is  16,  from  a  point  whose  distance  from  the  centre  is  17  ? 

12.  What  is  the  length  of  the  longest  straight  line  which  can  be 
drawn  on  a  floor  33  ft.  4  in.  long,  and  16  ft.  3  in.  wide  ? 

13.  A  ladder  32  ft.  6  in.  long  is  placed  so  that  it  just  reaches  a 
window  26  ft.  above  the  street ;  and  when  turned  about  its  foot,  just 
reaches  a  window  16  ft.  6  in.  above  the  street  on  the  other  side.  Find 
the  width  of  the  street. 

14.  The  altitude  of  an  equilateral  triangle  is  5  ;  what  is  its  side  ? 

15.  Find  the  length  of  the  diagonal  of  a  square  whose  side  is 
1  ft.  3  in. 

16.  One  of  the  non-parallel  sides  of  a  trapezoid  is  perpendicular 
to  the  bases.  If  the  length  of  this  side  is  40,  and  of  the  parallel  sides 
31  and  22,  respectively,  what  is  the  length  of  the  other  side  ? 

17.  The  length  of  the  tangent  to  a  circle,  whose  diameter  is  80, 
from  a  point  without  the  circumference,  is  42.  What  is  the  distance 
of  the  point  from  the  centre  ? 

18.  If  the  length  of  the  common  chord  of  two  intersecting  circles 
is  16,  and  their  radii  are  10  and  17,  what  is  the  distance  between 
their  centres  ?     (§  178.) 


144 


PLANE   GEOMETRY.  — BOOK  III. 


DEFINITIONS. 

275.  The  projection  of  a  point  upon  a  straight  line  of 
indefinite  length,  is  the  foot  of  the  per- 
pendicular drawn  from  the  point  to  the 
line. 

Thus,  if  line  AA^  be  perpendicular  to 
line  CD,  the  projection  of  point  A  on 
line  CD  is  point  A\ 

276.  The  projection  of  a  finite  straight  line  upon  a  straight 
line  of  indefinite  length,  is  that  portion  of  the  second  line 
included  between  the  projections  of  the  extremities  of  the 
first. 

Thus,  if  lines  AA  and  BB^  be  perpendicular  to  line  CD, 
the  projection  of  line  AB  upon  line  CD  is  line  AB\ 


Prop.  XXV.     Theorem 

277.  In  any  triangle,  the  square  of  the  side  opposite  an 
acute  angle  is  equal  to  the  sum  of  the  squares  of  the  other 
two  sides,  minus  tivice  the  product  of  one  of  these  sides  and^ 
the  projection  of  the  other  side  upon  it. 


Fig.  ^. 

Given  C  an  acute  Z  of  A  ABC,  and  CD  the  projection  of 
side  AC  upon  side  CB,  produced  if  necessary.  (§  276) 

To  Prove    AB^  ^BC" -\-  AC'  -2BCx  CD. 
Proof.     Draw  line  AD ;  then,  AD±  CD.  (§  276) 

There  will  be  two  cases  according  as  D  falls   on  CB 
(Fig.  1),  or  on  CB  produced  (Fig.  2). 


SIMILAK  POLYGONS. 


145 


In  Fig.  1,  BD  =  BC-  CD. 

In  Fig.  2,  BD=CD-  EG. 

Squaring  both  members  of  the  equation,  we  have  by  the 
algebraic  rule  for  the  square  of  the  difference  of  two  num- 
bers, in  either  case, 

Bff=^BG'+CD^-2BCxCD. 

Adding  AD"  to  both  members, 

Iff  +BD' ^W  +  Aff  +  W -2 BG  X  CD, 
But  in  rt.  A  ABD  and  ACD, 

Aff  +  Bff  =  A&, 

and  Aff  +  Cff  =  AG^.  (§  272) 

Substituting  these  values,  we  have 

AB'=BG^  +  AC^-2BCxGD. 


Prop.  XXVI.     Theorem. 

278.  In  any  triangle  having  an  obtuse  angle,  the  square  of 
the  side  opposite  the  obtuse  angle  is  equal  to  the  sum  of  the 
squares  of  the  other  two  sides,  plus  twice  the  product  of  one  of 
these  sides  and  the  projection  of  the  other  side  upon  it. 


Given  C  an  obtuse  Z  of  A  ABG,  and  GD  the  projection 
of  side  AG  upon  side  BG  produced. 

To  Prove     AB'  =  BG'  -{-  AG'  +  2  BG  x  GD. 

(We  have  BD  =  BG -I-  GD;  square  both  members,  using 
the  algebraic  rule  for  the  square  of  the  sum  of  two  num- 
bers, and  then  add  AD'  to  both  members.) 


146 


PLANE   GEOMETRY.— BOOK  III. 


Prop.  XXVII.     Theorem. 

279.  In  any  triangle^  if  a  median  he  drawn  from  the  vertex 
to  the  base, 

I.  The  sum  of  the  sqiiares  of  the  other  two  sides  is  equal 
to  twice  the  square  of  half  the  base,  plus  twice  the  square  of 
the  median. 

II.  The  difference  of  the  squares  of  the  other  two  sides  is 
equal  to  twice  the  product  of  the  base  and  the  projection  of  the 
median  upon  the  base. 

G 


Given  DE  the  projection  of  median  CD  upon  base  AB  of 
A  ABC',  s^nd  AC  >BC. 


To  Prove 


I.   AC  -\-BC=2AD'  +  2CD. 


II.   AC  -BC  =2ABxDE. 

Proof.     Since  AC  >  BC,  E  falls  between  B  and  D. 
Then,  Z  ADC  is  obtuse,  and  Z  BDC  acute. 
Hence,  in  A  ADC  and  BDC, 

AG^  =  AD'  +  Gff  +  2ADx  DE, 
and  BC^  =  BD"  +  Cff-2BDx  DE. 

But  by  hyp.,  BD  =  AD. 

.'.  AG""  =  Aff  +CD'-\-ABx  DE, 
and  W  =  ad'  +Gff-ABx  DE. 

Adding  (1)  and  (2),  we  have 

AG'  +  BC'  =  2AD'-^2GD\ 
Subtracting  (2)  from  (1),  we  have 

AG'  -BG'  =  2ABx  DE. 


(§  278) 
(§  277) 

(1) 
(2) 


SIMILAR  POLYGONS.  I47 

Prop.  XXVIII.     Theorem. 

280.  If  any  two  chords  be  drawn  through  a  fixed  point 
within  a  circle,  the  product  of  the  segments  of  one  chord  is 
equal  to  the  product  of  the  segments  of  the  other. 


Given  AB  and  A'B'  any  two  chords  passing  through  fixed 
point  P  within  O  AA'B. 
To  Prove  AP  x  BP=  A'P  x  B'P. 

Proof.     Draw  lines  AA'  and  BB'. 
Then,  in  A  AA'P  and  BB'P, 

ZA=ZB', 
for  each  is  measured  by  ^  arc  A'B.  (?) 

In  like  manner,  Z  A'  =  Z  B. 

Then,  A  AA'P  and  BB'P  are  similar.  (?) 

In  similar  A  AA'P  and  BB'P,  sides  AP  and  B'P  are 

homologous,  as  also  are  sides  A'P  and  BP.  (§  258) 

.-.  AP:A'P=B'P:BP.  (?) 

.-.  APxBP=  A'P  X  B'P.  (?) 

281.  Sch.     The  proportion  of  §  280  may  be  written 
AP_B^        AP_   1 
A'P~BP'  ^^  AJP~BP' 
B'P 

If  two  magnitudes,  such  as  the  segments  of  a  chord  pass- 
ing through  a  fixed  point,  are  so  related  that  the  ratio  of 
any  two  values  of  one  is  equal  to  the  reciprocal  of  the  ratio 
of  the  corresponding  values  of  the  other,  they  are  said  to 
be  reciprocally  proportional. 


148  PLANE   GEOMETRY.  — BOOK  III. 

Then,  the  theorem  may  be  written, 

If  any  two  chords  be  drawn  through  a  fixed  point  within 
a  cirdej  their  segments  are  reciprocally  proportional. 

Prop.  XXIX.     Theorem. 

282.  If  through  a  fixed  poirit  without  a  circle  a  secant  and 
a  tangent  be  drawn,  the  product  of  the  ivhole  secant  and  its 
external  segment  is  equal  to  the  square  of  the  tangent. 


Given  AP  a  secant,  and  CP  a  tangent,  passing  through 
fixed  point  P  without  O  ABC. 

To  Prove  APx  BP=  UP. 

{/.  A  =  Z.  BOP,  for  each  is  measured  by  ^  arc  BC  (?) ; 
then  A  AOP  and  BCP  are  similar,  and  their  homologous 
sides  are  proportional.) 

283.  Cor.  I.  If  through  a  fixed  point  without  a  circle  a 
secant  and  a  tangent  be  drawn,  the  tangent  is  a  mean  pro- 
portional between  the  whole  secant  and  its  external  segment. 

284.  Cor.  n.     If  any  two  secants         ^ — -^ 
be  drawn  through  a  fixed  point  without       /  ^^^^^^^^ 
a  circle,  the  product  of  one  and  its 
external  segment  is  equal  to  the  prod- 
uct of  the  other  and  its  external  seg- 
ment.                                                                  j^~ 

Given  P  any  point  without  OABC,  and  AP  and  A'P 
secants  intersecting  the  circumference  at  A  and  B,  and  A' 
and  B',  respectively. 


SIMILAR  POLYGONS.  149 

To  Prove  APxBP=  A'P  x  B'P. 

(We  have  AP  x  BP  and  A'P  x  B'P  each  equal  to  CP.) 

285.  Cor.  III.  If  any  two  secants  be  drawn  through  a 
fixed  point  without  a  circle,  the  whole  secants  and  their  ex- 
ternal segments  are  reciprocally  proportional  (§  281). 

EXERCISES. 

19.  Find  the  length  of  the  common  tangent  to  two  circles  whose 
radii  are  11  and  18,  if  the  distance  between  their  centres  is  25, 

20.  AB  is  the  hypotenuse  of  right  triangle  ABC.  If  perpendicu- 
lars be  drawn  to  AB  at  A  and  JB,  meeting  AC  produced  at  Z),  and  BC 
produced  at  E,  prove  triangles  ACE  and  BCD  similar. 


Pbop.  XXX.     Theorem. 

286.  In  any  triangle,  the  product  of  any  two  sides  is  equal 
to  the  diameter  of  the  circumscribed  circle,  multiplied  by  the 
perpendicular  drawn  to  the  third  side  from  the  vertex  of  the 
opposite  angle. 


Given  AD  a  diameter  of  the  circumscribed  QAOD  of 
A  ABC,  and  line  AE  ±  BC. 

To  Prove  AB  x  AC  =  AD  x  AE. 

(In  rt.  AABD  and  ACE,  ZD  =  ZC;  then,  the  A  are 
similar,  and  their  homologous  sides  are  proportional.) 

287.  Cor.  In  any  triangle,  the  diameter  of  the  circum- 
scribed circle  is  equal  to  the  product  of  any  two  sides  divided 
by  the  perpendicular  drawn  to  the  third  side  from  the  vertex 
of  the  opposite  angle. 


150 


PLANE   GEOMETRY.  — BOOK  III. 


Prop.  XXXI.     Theorem. 

288.  In  any  triangle,  the  product  of  any  two  sides  is 
equal  to  the  product  of  the  segments  of  the  third  side  formed 
by  the  bisector  of  the  opposite  angle,  plus  the  square  of  the 
bisector. 


Tic 


— t.— - 
E 


Given,  in   A  ABC,  line  AD  bisecting  /.A,  meeting  BQ 
't  D. 

To  Prove       AB  x  AC  =BD  x  DC -\-  Ajf. 
Proof.     Circumscribe  a  O  about  A  ABC\  produce  AD  to 
meet  the  circumference  at  E,  and  draw  line  CE. 
Then  in  A  ABD  and  ACE,  by  hyp., 

Z  BAD  =  Z  CAE. 
Also,  ZB  =  ZE, 

since  each  is  measured  by  i  arc  AC.  (?) 

Then,  A  ABD  and  ACE  are  similar.  (?) 

Tn  A  ABD  and  ACE,  sides  AB  ?,nd  AE  are  homologous, 

(§  258) 


as  also  are  sides  AD  and  AC. 

.'.  AB:AD  =  AE:  AC. 
.-.  ABxAC=ADxAE 

=  ADx  (DE  -h  AD) 
=  AD  X  DE  +  Alf. 
But  ADx  DE  =  BD  X  DC. 

.'.  AB  X  AC=BD  X  DC  +  Aff. 


(?) 


(§  280) 


SIMILAR  POLYGONS.  151 


EXERCISES. 

21.  The  square  of  the  altitude  of  an  equilateral  triangle  is  equal 
to  three-fourths  the  square  of  the  side. 

22.  If  AD  is  the  perpendicular  from  A  to  side  BC  of  triangle 
ABC,  prove  _        

AB^-AC^=  BD^-  CD^. 

23.  If  one  leg  of  a  right  triangle  is  double  the  other,  the  perpen- 
dicular from  the  vertex  of  the  right  angle  to  the  hypotenuse  divides 
it  into  segments  which  are  to  each  other  as  1  to  4.     (§  271.) 

24.  If  two  parallels  to  side  BC  of  triangle  ABC  meet  sides  AB 
and  ^C  at  D  and  F,  and  E  and  G,  respectively,  prove 

BD^BF=DF^     (§247.) 
CE     CG     EG      ^^        ^ 

25.  C  and  D  are  respectively  the  middle  points  of  a  chord  AB 
and  its  subtended  arc.  If  AD  =  12  and  CD  =  8,  what  is  the  diame- 
ter of  the  circle  ?     (§  27L) 

26.  If  AD  and  BE  are  the  perpendiculars  from  vertices  A  and  B 
of  triangle  ABC  to  the  opposite  sides,  prove 

AC  :  DC  =  BC  :  EC. 
(Prove  AACD  and  BCE  similar.) 

27.  If  D  is  the  middle  point  of  side  BC  of  triangle  ABCj  right- 
angled  at  C,  prove  AE^  -  A&  =  3  CD^. 

28.  The  diameters  of  two  concentric  circles  are  14  and  50  units, 
respectively.  Find  the  length  of  a  chord  of  the  greater  circle  which 
is  tangent  to  the  smaller.     (§  273.) 

29.  The  length  of  a  tangent  to  a  circle  from  a  point  8  units  dis- 
tant from  the  nearest  point  of  the  circumference,  is  12  units.  What 
is  the  diameter  of  the  circle  ? 

(Let  X  represent  the  radius. ) 

30.  The  non-parallel  sides  AD  and  BC  ot  trapezoid  ABCD  inter- 
sect at  0.  If  AB  =  15,  CD  =  24,  and  the  altitude  of  the  trapezoid  is 
8,  what  is  the  altitude  of  triangle  GAB  ?     (§  2G4.) 

(Draw  CEWAD.) 

31.  If  the  equal  sides  of  an  isosceles  right  triangle  are  each  18 
units  in  length,  what  is  the  length  of  the  median  drawn  from  the 
vertex  of  the  right  angle  ? 

32.  The  non-parallel  sides  of  a  trapezoid  are  each  53  units  in 
length,  and  one  of  the  parallel  sides  is  50  units  longer  than  the  other. 
Find  the  altitude  of  the  trapezoid. 


152  PLANE   GEOMETRY.  —  BOOK  III. 

33.  AB  is  a  chord  of  a  circle,  and  CE  is  any  chord  drawn  through 
the  middle  point  C  of  arc  AB,  cutting  chord  AB  at  Z>.  Prove  AG 
a  mean  proportional  between  CD  and  CE. 

(Prove  AACD  and  ACE  similar.) 

34.  Two  secants  are  drawn  to  a  circle  from  an  outside  point.  If 
their  external  segments  are  12  and  9,  respectively,  while  the  internal 
segment  of  the  former  is  8,  what  is  the  internal  segment  of  the  latter  ? 
(§284.) 

35.  If,  in  triangle  ABC,  ZC=  120°,  prove 

AB^  =  BG^  +  AG^  -{-ACx  BC. 
(Fig.  of  Prop.  XXVI.     A  AC D  is  one-half  an  equilateral  A.) 

36.  BC  is  the  base  of  an  isosceles  triangle  ABC  inscribed  in  a 
circle.    If  a  chord  AD  be  drawn  cutting  BC  2it  E,  prove 

AD:AB  =  AB:  AE. 
(Prove  A  ABD  and  ABE  similar.) 

37.  Two  parallel  chords  on  opposite  sides  of  the  centre  of  a  circle 
are  48  units  and  14  units  long,  respectively,  and  the  distance  between 
their  middle  points  is  31  units.     What  is  the  diameter  of  the  circle  ? 

(Let  X  represent  the  distance  from  the  centre  to  the  middle  point 
of  one  chord,  and  31  —  «  the  distance  from  the  centre  to  the  middle 
point  of  the  other.  Then  the  square  of  the  radius  may  be  expressed 
in  two  ways  in  terms  of  x. ) 

38.  ABC  is  a  triangle  inscribed  in  a  circle.  Another  circle  is 
drawn  tangent  to  the  first-  externally  at  G,  and  AG  and  BC  are  pro- 
duced to  meet  its  circumference  at  D  and  E,  respectively.  Prove  tri- 
angles ABC  and  CDE  similar.     (§  197.) 

(Draw  a  common  tangent  to  the  (D  at  G.  Then  BC  and  CE  are 
arcs  of  the  same  number  of  degrees. ) 

39.  ABC  and  A'BG  are  triangles  whose 
vertices  A  and  A'  lie  in  a  parallel  to  their  com- 
mon base  BC.  If  a  parallel  to  BC  cuts  AB 
and  AG  Sit  D  and  E,  and  A'B  and  A'C  at  D' 
and  E',  respectively,  prove  DE  =  D'E'. 

(prove  ^=^^.) 
\  BC      BG   I 

40.  A  line  parallel  to  the  bases  of  a  trapezoid,  passing  through 
the  intersection  of  the  diagonals,  and  terminating  in  the  non-parallel 
sides,  is  bisected  by  the  diagonals.     (Ex.  39.) 

41.  If  the  sides  of  triangle  ABC  are  AB  =  10,  BG=U,  and 
CA  =  16,  find  the  lengths  of  the  three  medians.     (§  279,  I.) 


SIMILAR  POLYGONS.  I53 

42.  If  the  sides  of  a  triangle  are  AB  =  i,  AC=5,  and  BC  =  6, 
find  the  length  of  the  bisector  of  angle  A.     (§§  249,  288  ) 

P 

43.  The  tangents  to  two  intersecting  circles 
from  any  point  in  their  common  chord  produced 
are  equal.     (§  282.) 


44.  If  two  circles  intersect,  their  common  chord  produced  bisects 
their  common  tangents. 

45.  AB  and  A  C  are  the  tangents  to  a  circle  from  point  A.  If  CD 
be  drawn  perpendicular  to  radius  OB  at  D,  prove 

AB:OB  =  BD:  CD. 

(Prove  A  OAB  and  BCD  similar  by  §  262.) 

46.  ABC  is  a  triangle  inscribed  in  a  circle,  A  line  AD  is  drawn 
from  A  to  any  point  of  BC,  and  a  chord  BE  is  drawn,  making 
ZABE=ZADC.     Prove 

ABxAC  =  ADxAE. 

(Prove  AB  :  AE  =z  AD  :  AC.) 

47.  The  radius  of  a  circle  is  22^  units.  Find  the  length  of  a  chord 
which  joins  the  points  of  contact  of  two  tangents,  each  30  units  in 
length,  drawn  to  the  circle  from  a  point  without  the  circumference. 

(By  §  271,  2,  the  radius  is  a  mean  proportional  between  the  dis- 
tances from  the  centre  to  the  chord  and  to  the  point  without  the  cir- 
cumference ;  in  this  way  the  distance  from  the  centre  to  the  chord 
can  be  found.) 

48.  If,  in  right  triangle  ABC,  acute  angle  B  is  double  acute  angle 
A,  prove  AC^  =  3  BC\     (Ex.  104,  p.  71.) 

49.  Find  the  product  of  the  segments  of  any  chord  drawn  through 
a  point  9  units  from  the  centre  of  a  circle  whose  diameter  is  24  units. 

50.  The  hypotenuse  of  a  right  triangle  is  5,  and  the  perpendicular 
to  it  from  the  opposite  vertex  is  2f.  Find  the  legs,  and  the  segments 
into  which  the  perpendicular  divides  the  hypotenuse.     (§  271.) 

(Let  X  represent  one  of  the  segments  of  the  hypotenuse.) 

51.  State  and  prove  the  converse  of  Prop.  XIII. 

(Fig.  of  Prop.  XIII.  To  prove  Z  BAD  =  Z  CAD.  Produce  CA  to 
E,  making  AE  =  AB.) 

52.  State  and  prove  the  converse  of  Prop.  XIV. 
(Fig.  of  Prop.  XIV.    Lay  off  AF  =  AB.) 


154  PLANE  GEOMETEY.  — BOOK  III. 

53.  If  D  is  the  middle  point  of  hypotenuse  AB  of  right  triangle 
ABC^  prove 

(JD^  =  \  {AB^  +  BC^  +  CA^) .     (Ex.  83,  p.  69. ) 

54.  If  a  line  be  drawn  from  vertex  C  of  isosceles  triangle  ABC^ 
meeting  base  AB  produced  at  Z>,  prove 

CD^  -  CB^  =  ADx  BD.     (§  278.) 

55.  If  AB  is  the  base  of  isosceles  triangle  ABC^  and  AD  be  drawn 
perpendicular  to  BC^  prove 

3  AL^  +  Blf  +  2CD^  =  AB^  +  BC^  +  (Ja\ 
(We  have  3  AB^  =  AD^  +  2  AD^. ) 

56.  The  middle  points  of  two  chords  are  distant  5  and  9  units, 
respectively,  from  the  middle  points  of  their  subtended  arcs.  If  the 
length  of  the  first  chord  is  20  units,  find  the  length  of  the  second. 

(Find  the  diameter  by  aid  of  §  270,  1.) 

57.  The  sides  AB  and  AC,  of  triangle  ABC,  are  16  and  9,  respec- 
tively, and  the  length  of  the  median  drawn  from  C  is  11.  Find  side 
BC.     (§279,1.) 

58.  The  diameter  which  bisects  a  chord  whose  length  is  33|  units, 
is  35  units  in  length.  Find  the  distance  from  either  extremity  of  the 
chord  to  the  extremities  of  the  diameter. 

(Let  X  represent  one  segment  of  the  diameter  made  by  the  chord.) 

59.  The  equal  angles  of  an  isosceles  triangle  are  each  30°,  and  the 
equal  sides  are  each  8  units  in  length.  What  is  the  length  of  the 
base  ?     (Ex.  104,  p.  71.) 

60.  The  diagonals  of  a  trapezoid,  whose  bases  are  AD  and  BC, 
intersect  at  U.  If  AE  =  9,  EC  =  3,  and  BD  =  16,  find  BE  and 
ED. 

(A  AED  and  BEC  are  similar.     Find  BE  by  §  237. ) 

61.  Prove  the  theorem  of  §  284  by  drawing  A'B  and  AB'. 

62.  The  parallel  sides,  AD  and  BC,  of  a  circumscribed  trapezoid 
are  18  and  6,  respectively,  and  the  other  two  sides  are  equal  to  each 
other.     Find  the  diameter  of  the  circle. 

(Find  AB  by  Ex.  31,  p.  100.    Draw  through  5  a  ||  to  CD.) 

63.  An  angle  of  a  triangle  is  acute,  right,  or  obtuse  according  as 
the  square  of  the  opposite  side  is  less  than,  equal  to,  or  greater  than, 
the  sum  of  the  squares  of  the  other  two  sides. 

(Prove  by  Beductio  ad  Ahsurdum. ) 

64.  Is  the  greatest  angle  of  a  triangle  whose  sides  are  3,  5,  and  6, 
acute,  right,  or  obtuse  ? 


SIMILAR  POLYGONS.  I55 

65.  Is  the  greatest  angle  of  a  triangle  whose  sides  are  8,  9,  and  12, 
acute,  right,  or  obtuse  ? 

66.  Is  the  greatest  angle  of  a  triangle  whose  sides  are  12,  35,  and 
37,  acute,  right,  or  obtase  ? 

67.  If  two  adjacent  sides  and  one  of  the  diagonals  of  a  parallelo- 
gram are  7,  9,  and  8,  respectively,  find  the  other  diagonal. 

(One-half  of  either  diagonal  is  a  median  of  the  A  whose  sides  are, 
respectively,  the  given  sides  and  the  other  diagonal  of  the  O.) 

68.  If  D  is  the  intersection  of  the  perpendiculars  from  the  vertices 
of  triangle  ABC  io  the  opposite  sides,  prove 

AB^  -  AG^  =  Bjf  -  CD^.     (§  272.) 

69.  If  a  parallel  to  hypotenuse  AB  of  right  triangle  ABC  meets 
AC  and  BC  a.t  D  and  E,  respectively,  prove 

AE^  +  B&  =  AB^  +  DE"^. 

70.  The  diameters  of  two  circles  are  12  and  28,  respectively,  and 
the  distance  between  their  centres  is  29.  Find  the  length  of  the 
common  tangent  which  cuts  the  straight  line  joining  the  centres. 

(Find  the  ±  drawn  from  the  centre  of  the  smaller  O  to  the  radius 
of  the  greater  O  produced  through  the  point  of  contact.) 

71.  State  and  prove  the  converse  of  Prop.  XXIII. ,  III. 
(Fig.  of  Prop.  XXIII.     A  ABC  and  ACD  are  similar.) 

72.  State  and  prove  the  converse  of  Prop.  XXIIL,  II. 

73.  The  sum  of  the  squares  of  the  distances  of 
any  point  in  the  circumference  of  a  circle  from 
the  vertices  of  an  inscribed  square,  is  equal  to 
twice  the  square  of  the  diameter  of  the  circle. 
(§195.)  ______         _,       • 

(To  prove  P^^+PB^-fP(7^+PD^=2^C^.)  m  ^^    ^D 

74.  The  sides  AB,  BC,  and  GA,  of  triangle  ABC,  are  13,  14,  and 
15,  respectively.  Find  the  segments  into  which  AB  and  BC  are  di- 
vided by  perpendiculars  drawn  from  C  and  A,  respectively. 

{A  BAC  and  A  CB  are  acute  by  §  98.    Find  the  segments  by  §  277.) 

75.  In  right  triangle  ABC  is  inscribed  a  square  DEFGr,  having 
its  vertices  D  and-  C  in  hypotenuse  BC,  and  its  vertices  E  and  F 
in  sides  AB  and  AC,  respectively.    Prove  BD  :  DE  =  DE  :  CQ. 

(Prove  A  BDE  and  CFG  similar.) 

Note.    For  additional  exercises  on  Book  III.,  see  p.  226. 


156  PLANE   GEOMETKY.— BOOK   III. 

CONSTRUCTIONS. 

Prop.  XXXII.     Problem. 

289.    To  divide  a  given  straight  line  into  any  number  of 
equal  parts. 

.^ 

_  5<     \   \ 


A         H         K         L  B 

Given  line  AB. 
Required  to  divide  AB  into  four  equal  parts. 

Construction.  On  the  indefinite  line  AC,  take  any  con- 
venient length  AD.,  on  DC  take  DE  —  AD\  on  EG  take 
EF=AD',  on  FC  take  FG  =  AD  ;  and  draw  line  BG. 

Draw  lines  DH,  EK,  and  FL  II  BG,  meeting  AB  at  H, 
Kf  and  L,  respectively. 

.'.  AH=  HK=  KL  =  LB.  (§  242) 

Prop.  XXXIII.     Problem. 

290.  To  construct  a  fourth  proportional  (§  231)  to  three 
given  straight  lines. 


^^-■' 


my 
A -"---■ 


P        F  Q 

Given  lines  m,  n,  and  p. 

Required  to  construct  a  fourth  proportional  to  m,  n,  and  p. 
Construction.      Draw  the  indefinite  lines  AB  and  AC, 
making  any  convenient  /.  with  each  other. 


SIMILAR  POLYGONS.  X67 

On  AB  take  AD  =  m ;  on  DB  take  DE  =  n;  on  AC  take 
AF  =  p. 

Draw  line  DF;  also,  line  EG  II  DF,  meeting  ^Oat  G. 
Then,  FG  is  a  fourth  proportional  to  ?ai,  71,  and  2?- 
Proof.     Since  DF  is  11  to  side  EG  of  A  ^^G^, 

AD:DE  =  AF:FG.  (?) 

That  is,  m:n=p:  FG. 

291.  Cor.    If  we  take  AF  =  n,  the  proportion  becomes 

m:n  =  n:  FG. 
In  this  case,  FG  is  a  ^/uVd  proportional  (§  230)  to  m  and  n. 

Prop.  XXXIV.     Problem. 

292.  To  construct  a  mean  proportional  (§  230)  between  two 
given  straight  lines. 

P 


A  rn  B     n     G      E 

Given  lines  m  and  n. 

Required  to  construct  a  mean  proportional  between  m 
and  71. 

Construction.  On  the  indefinite  line  AE,  take  AB  =  m ; 
on  BE  take  BC  =  n. 

With  J.(7  as  a  diameter,  describe  the  semi-circumference 
ADC. 

Draw  line  BD1.AC,  meeting  the  arc  at  D. 

Then,  BD  is  a  mean  proportional  between  m  and  n. 

(The  proof  is  left  to  the  pupil ;  see  §  270.) 

293.  Sch.  By  aid  of  §  292,  a  line  may  be  constructed 
equal  to  Va,  where  a  is  any  number  whatever. 

Thus,  to  construct  a  line  equal  to  V3,  we  take  AB  equal 
to  3  units,  and  BC  equal  to  1  unit.         

Then,  BD  =  VABxBC  (§  232)  =  V3  x  1  =  V3. 


158  PLANE  GEOMETRY.— BOOK  IH. 


Prop.  XXXV.     Problem. 

294.    To  divide  a  given  straight  line  into  parts  proportional 
to  any  number  of  given  lines. 

F  -C 


^-"'     /        /  / 

/ 


G       H 
Given  line  AB,  and  lines  m,  n,  and  p. 

Required  to  divide  AB  into  parts  proportional  to  m,  n, 
and  p. 

Construction.  On  the  indefinite  line  AG,  take  AD  =  m ; 
on  DC  take  DE  =  n;  on  EG  take  EF  =  p\  and  draw  line 
BF. 

Draw  lines  DG  and  EH  II  to  ^l^",  meeting  AB  at  G^  and 
IT,  respectively. 

Then,  AB  is  divided  into  parts  AG,  GH,  and  HB  propor- 
tional to  m,  ti,  and  p,  respectively. 

Proof.     Since  DGi^W  to  side  ^iT  of  A  AEH, 

AH^AG^GH  .^. 

AE     AD     DE'  ^'^ 

rri-i    ,  •  AH     AG      GH  /^x 

That  IS,  — — -  = = (1) 

'  AE       m         n  ^  ^ 

And  since  EH  is  II  to  side  BF  of  A  ABF, 

AH^HB^HB  ,2^ 

AE     EF       p  '  ^  ^ 

From  (1)  and  (2),  4^  =  ^=^.  (?) 

m         n         p 


Ex.  76.   Construct  a  Ime  equal  to  V2 ;  to  V5  ;  to  \/6. 


SIMILAR  POLYGONS. 


Prop.  XXXVI.     Problem. 


15^ 


295.  Upon  a  given  side,  homologous  to  a  given  side  of  a 
given  polygon,  to  construct  a  polygon  similar  to  the  given 
polygon. 

D 


A  B  A'  B' 

Given  polygon  ABODE,  and  line  AB\ 

Required  to  construct  upon  side  A'B',  homologous  to  AB, 
a  polygon  similar  to  ABODE. 

Construction.  Divide  polygon  ABODE  into  A  by  draw- 
ing diagonals  EB  and  EO. 

At  A  construct  ZB'A'E'==Z.A',  and  draw  line  B'E', 
making  Z  A  B'E'  =  ZABE,  meeting  A'E'  at  E'. 

Then,  A  A' B'E'  will  be  similar  to  A  ABE.  (?) 

In  like  manner,  construct  AB'O'E'  similar  to  ABOE, 
and  A  O'D'E'  similar  to  A  ODE. 

Then,  polygon  A' B' O'D'E'  will  be  similar  to  polygon 
ABODE.  (§  266) 

296.  Def.  A  straight  line  is  said  to  be  divided  by  a 
given  point  in  extreme  and  mean  ratio  when  one  of  the  seg- 
ments (§  250)  is  a  mean  proportional  between  the  whole 
line  and  the  other  segment. 

D  A  C       B 


Thus,  line  AB  is  divided  internally  in  extreme  and  mean 
ratio  at  0  if 

AB:AO=AO:BO', 

and  externally  in  extreme  and  mean  ratio  at  D  if 
AB'.AD  =  AD'.  BD. 


PLANE  GEOMETRY.  — BOOK  m. 


To 


Prop.  XXXVII.     Problem. 
divide  a  given  straight  line  in  extreme  and  mean 


ratu 


'296). 


Given  line  AB. 

Required  to  divide  AB  in  extreme  and  mean  ratio. 
Construction.     Draw  line  BE  ±  AB,  and  equal  to  i  AB. 
With  ^  as  a  centre  and  EB  as  a  radius,  describe  O  BFG. 
Draw  line  AE  cutting  the  circumference  at  F  and  G. 
On  AB  take  AC  =  AF;  on  BA  produced,  take  AD  =  AG. 
Then,  AB  is  divided  at  O  internally,  and  at  D  externally, 

in  extreme  and  mean  ratio. 

• 

Proof.     Since  AG  is  a  secant,  and  AB  a  tangent, 

AG:AB  =  AB:  AF.  (§  283) 

.-.  AG:AB  =  AB'.AG.  (1) 

.-.  AG-AB:AB  =  AB-AC:AG.  (?) 

.-.  AB:AG-AB  =  ACiBC.  (?) 

But  by  cons.,  AB  =2BE  =  FG.  (2) 

.-.  AG-AB  =  AG-FG  =  AF=  AC. 
Substituting,     AB :  AC  =  AC :  BO.  (3) 

Therefore,  AB  is  divided  at  C  internally  in  extreme  and 
mean  ratio. 

Again,  from  (1), 

AG  +  AB  :  AG  =  AB-{-  AC :  AB.  (?) 

But,  AG-^AB^AD  +  AB  =  BD. 

And  by  (2),     AB -\- AC  =  FG  +  AF  ==  AG. 


SIMILAR  POLYGONS.  161 

„•.  BD:AG  =  AG:AB. 

.'.  AB :  AG  =  AG :  BD.  '  (?) 

.-.  AB:AD==AD:BD. 

Therefore,  AB  is  divided  at  D  externally  in  extreme  and 
mean  ratio. 

298.  Cor.    If  AB  be  denoted  by  m,  and  AG  by  x,  propor- 
tion (3)  of  §  297  becomes 

m:x  =  x:m  —  x. 

.'.  x^  =  m  (m  -x)  =  m^-  mx.     (§  232) 
Or,  a?  +  Ttix  =  m^. 

Multiplying  by  4,  and  adding  m^  to  both  members, 

4  ic^  +  4  mx  +  m^  =  4  m^  +  m^  =  5  m^ 
Extracting  the  square  root  of  both  members, 

2x-\-  m  =  ±  mV5. 
Since  x  cannot  be  negative,  we  take  the  positive  sign 
before  the  radical  sign ;  then, 

2x=  mV5  —  wi. 


EXERCISES. 

77.  To  inscribe  in  a  given  circle  a  triangle  similar  to  a  given 
triangle.     (§  261.) 

(Circumscribe  a  O  about  the  given  A,  and  draw  radii  to  the 
vertices.) 

78.  To  circumscribe  about  a  given  circle  a  triangle  similar  to  a 
given  triangle.     (§  262.) 


Book  IV. 


AREAS    OF    POLYGONS 
Prop.  I.     Theorem. 

299.  Two  rectangles  having  equal  altitudes  are  to  each 
other  as  their  bases. 

Note.  The  words  "rectangle,"  "parallelogram,"  "triaugle,"  etc., 
in  the  propositions  of  Book  IV.,  mean  the  amount  of  surface  in  the 
rectangle,  parallelogram,  triangle,  etc. 

Case  I.      When  the  bases  are  commensurable. 

B .a  F, , ,G 


E 


A        K 
Given  rectangles  ABCD  and  EFGH,  with  equal  altitudes 


AB  and  EF,  and  commensurable  bases  AD  and  EH. 

TnProv«  ABCD  ^  AD  ^ 

EH 
Proof. 


EFGH 
Let  AK  be  a  common  measure  of  AD  and  EH, 
and  let  it  be  contained  5  times  in  AD,  and  3  times  in  EH. 

.-.  ^  =  1  (1) 

EH     3  ^  ^ 

Drawing  Js  to  AD  and  EH  through  the  several  points  of 

division,  rect.  ABCD  will  be  divided  into  5  parts,  and  rect. 

EFGH  into  3  parts,  all  of  which  parts  are  equal.       (§  114) 

ABCD  ^  5 

EFGH     3 

ABCD  ^  AD 

EFGH     EH 

162 


From  (1)  and  (2), 


(2) 
(?) 


AREAS  OF  POLYGONS.  163 

Case  II.     When  the  bases  are  incommensurable. 


"  "  K    ^ 

Given  rectangles  ABCD  and  EFGH,  with  equal  altitudes 
AB  and  EF,  and  incommensurable  bases  AD  and  EH. 

To  Prove  ABCD  ^  AD  ^ 

EFGH     EH 

Proof.  Divide  AD  into  any  number  of  equal  parts,  and 
apply  one  of  these  parts  to  EH  as  a  unit  of  measure. 

Since  AD  and  EH  are  incommensurable,  a  certain  num- 
ber of  the  parts  will  extend  from  E  to  K,  leaving  a  re- 
mainder KH  <  one  of  the  equal  parts. 

Draw  line  KL  A.  EH,  meeting  FG  at  L. 

Then,  since  AD  and  EK  are  commensurable, 

4^^  =  4R.  (§  299,  Case  I.) 

EFLK     EK  ^         '  ^ 

Now  let  the  number  of  subdivisions  of  AD  be  indefinitely 
increased. 

Then  the  unit  of  measure  will  be  indefinitely  diminished, 
and  the  remainder  KH  will  approach  the  limit  0. 

Then,  will  approach  the  limit  ^^^^^> 

and  will  approach  the  limit  — —  • 

EK  ^^  EH 

By  the  Theorem  of  Limits,  these  limits  are  equal.         (?) 

ABCD  ^  AD 

■  ■  EFGH     EH 

300.  Cor.  Since  either  side  of  a  rectangle  may  be  taken 
as  the  base,  it  follows  that 

Two  rectangles  having  equal  bases  are  to  each  other  as  their 
altitudes. 


161 


PLANE  GEOMETRY.  — BOOK  IV. 


Prop.  II.     Theorem. 

301.   Any  two  rectangles  are  to  each  other  as  the  products 
of  their  bases  by  their  altitudes. 


M 


a\ 


b  b'  b' 

Given  M  and  N  rectangles,  with  altitudes  a  and  a\  and 
bases  b  and  b\  respectively. 

M  _  a  xb 

]sr~ 


To  Prove 


a'xb' 

Proof.   Let  Rhe  a.  rect.  with  altitude  a  and  base  b'. 
Then,  since  rectangles  M  and  E  have  equal  altitudes,  they 
are  to  each  other  as  their  bases.  (§  299) 

.-.  ^=L  (1) 

R      b'  ^  ^ 

And  since  rectangles  R  and  ^  have  equal  bases,  they  are 

to  each  other  as  their  altitudes.  (?) 

R^a 

"  J^     a'' 

Multiplying  (1)  and  (2),  we  have 

MR         M^  axb  ^ 

R      N'        M     a'  xb' 


(2) 


DEFINITIONS. 

302.  The  area  of  a  surface  is  its  ratio  to  another  surface, 
called  the  unit  of  surface,  adopted  arbitrarily  as  the  unit  of 
measure  (§  180). 

The  usual  unit  of  surface  is  a  square  whose  side  is  some 
linear  unit;  for  example,  a  square  inch  or  a  square  foot. 

303.  Two  surfaces  are  said  to  be  equivalent  (=o=),  when 
their  areas  are  equal. 


AREAS  OF  POLYGONS. 


165 


304.  The  dimensions  of  a  rectangle  are  its  base  and 
altitude. 

Prop.  III.     Theorem. 

305.  The  area  of  a  rectangle  is  equal  to  the  product  of  its 
base  and  altitude. 

Note.     In  all  propositions  relating  to  areas,  the  unit  of  surface 
(§  302)  is  understood  to  be  a  square  whose  side  is  the  linear  unit. 


N 


Given  a  and  h,  the  altitude  and  base,  respectively,  of 
rect.  3f ;  and  N  the  unit  of  surface,  i.e.,  a  square  whose 
side  is  the  linear  unit. 

To  Prove  that,  if  N  is  the  unit  of  surface, 
area  M=  a  xb. 

Proof.  Since  any  two  rectangles  are  to  each  other  as  the 
products  of  their  bases  by  their  altitudes  (§  301), 

N     1x1 
But  since  N  is  the  unit  of  surface,  the  ratio  of  il!f  to  JV"  is 
the  area  of  M.  (§  302) 

.-.  area  Jtf=  a  x  b. 

306.  Sch.  I.  The  statement  of  Prop.  III.  is  an  abbrevia- 
tion of  the  following : 

If  the  unit  of  surface  is  a  square  whose  side  is  the  linear 
unit,  the  number  which  expresses  the  area  of  a  rectangle  is 
equal  to  the  product  of  the  numbers  which  express  the 
lengths  of  its  sides. 

An  interpretation  of  this  form  is  always  understood  in 
every  proposition  relating  to  areas. 


166 


PLANE  GEOMETRY.  — BOOK  IV. 


307.   Cor.     The  area  of  a  square  is  equal  to  the  square  of 
its  side. 


308.  Sch.  II.  If  the  sides  of  a  rec- 
tangle are  multiples  of  the  linear  unit,  the 
truth  of  Prop.  III.  may  be  seen  by  dividing 
the  figure  into  squares,  each  equal  to  the 
unit  of  surface. 

Thus,  if  the  altitude  of  rectangle  A  is 
5  units,  and  its  base  6  units,  the  figure  can  be  divided  into 
30  squares. 

In  this  case,  30,  the  number  which  expresses  the  area  of 
the  rectangle,  is  the  product  of  6  and  5,  the  numbers  which 
express  the  lengths  of  the  sides. 

Prop.  IV.     Theorem. 

309.  The  area  of  a  parallelogram  is  equal  to  the  product 
of  its  base  and  altitude. 

E       B      F        G 


A.  b  D 

Given  O  ABCD,  with  its  altitude  DF=  a,  and  its  base 
AD  =  h. 

To  Prove  area  ABCD  =  axb. 

Proof.     Draw  line  AE  II  DF,  meeting  CB  produced  at  E. 
Then,  AEFD  is  a  rectangle.  (?) 

In  rt.  A  ABE  and  DOF, 

AB  =  DC,  and  AE  =  DF.  (?) 

.-.  AABE  =  ADCF.  (?) 

Now  if  from  the  entire  figure  ADOE  we  take  A  ABE, 

there  remains   O  ABCD\   and  if  we   take  AZ)(7i^,  there 

remains  rect.  AEFD. 

.-.  area  ABCD  =  area  AEFD  =axb.        (§  305) 


AREAS  OF  POLYGONS.  ^q^ 

310.  Cor.  I.     Two  parallelograms  having  equal  bases  and 
equal  altitudes  are  equivalent  (§  303). 

311.  Cor.  II.     1.   Two  parallelograms  having  equal  alti- 
tudes are  to  each  other  as  their  bases. 

2.  Two  parallelograms  having  equal  bases  are  to  each  other 
as  their  altitudes. 

3.  Any  two  parallelograms  are  to  each  other  as  the  products 
of  their  bases  by  their  altitudes. 

Prop.  V.     Theorem. 

312.  The  area  of  a  triangle  is  equal  to  one-half  the  product 
of  its  base  and  altitude. 


B         E         h  C 

■   Given  A  ABC,  with  its   altitude  AE  =  a,  and  its  base 
BC=b. 

To  Prove  area  ABC  =  ^axb. 

(By  §  108,  AC  divides  CJ  ABCD  into  two  equal  A.) 


313.  Cor.  I.     Two  triangles  having  equal  bases  and  equal 
altitudes  are  equivalent. 

314.  Cor.  II.     1.   Two  triangles  having  equal  altitudes  are 
to  each  other  as  their  bases. 

2.  Two  triangles  having  equal  bases  are  to  each  other  as 
their  altitudes. 

3.  Any  two  triangles  are  to  each  other  as  the  products  of 
their  bases  by  their  altitudes. 

315.  Cor.  III.     A  triangle  is  equivaleyit  to  one-half  of  a 
parallelogram  having  the  same  base  and  altitude. 


168        PLANE  GEOMETEY.  — BOOK  IV. 


Prop.  VI.  Theorem. 

316.    The  area  of  a  trapezoid  is  equal  to  one-half  the  sum 
of  its  bases  multiplied  by  its  altitude. 


A         E  b  B 

Given  trapezoid  ABCD,  with  its  altitude  DE  equal  to  a, 
and  its  bases  AB  and  DC  equal  to  b  and  6',  respectively. 
To  Prove  area  ABCD  =  a  x  ^  (6  +  &'). 

(The  trapezoid  is  composed  of  two  A  whose  altitude  is  a, 
and  bases  b  and  V,  respectively.) 

317.  Cor.  Since  the  line  joining  the  middle  points  of 
the  non-parallel  sides  of  a  trapezoid  is  equal  to  one-half  the 
sum  of  the  bases  (§  132),  it  follows  that 

The  area  of  a  trapezoid  is  equal  to  the  product  of  its  alti- 
tude by  the  line  joining  the  middle  points  of  its  non-parallel 
sides. 

318.  Sch.  The  area  of  any  polygon  may  be  obtained  by 
finding  the  sum  of  the  areas  of  the  triangles  into  which  the 
polygon  may  be  divided  by  drawing  diagonals  from  any 
one  of  its  vertices. 

But  in  practice  it  is  better  to  draw  the  /^^""^--^ 

longest  diagonal,  and  draw  perpendicu-        /  \  \\ 

lars  to  it  from  the  remaining  vertices  of  /„_.].._ _  ]  \ 

the  polygon.     The  polygon  will  then  be  X.  j        'y^ 

divided  into  right   triangles  and  trape-  ^\[/^ 

zoids;    and  by  measuring  the  lengths  of 

the  perpendiculars,  and  of   the   portions  of  the   diagonal 

which   they   intercept,   the  areas   of  the  figures   may  be 
found  by  §§  312  and  316. 


i 


AREAS  OF  POLYGONS.  \QQ 


Prop.  VII.     Theorem. 

319.    Two  similar  triangles  are  to  each  other  as  the  squares 
of  their  homologous  sides. 
C 


Oiven  AB  and  A'B'  homologous  sides  of  similar  A  ABO 
and  A'B'C,  respectively. 

rn.  -o ABC  AB" 


*''      "                      A'B'C     A'B'' 

Proof.     Draw  altitudes  CD  and  CD'. 

.     ABC        AB  X  CD 
' '  A'B'C     A'B'  X  CD' 

(§ 

314,  3) 

AB  ^  CD 
A'B'  '^  CD' 

(1) 

J..                                CD  _  AB 
^""^^                              CD'- A'B'' 

(§  264) 

Substituting  this  value  in  (1), 

ABC  _  AB  ,,  AB  _ 

AB' 

A'B'C     A'B'     A'B'     A!B'' 

320.   Sch.     Two  similar  triangles  are  to  each  other  as  the 
squares  of  any  two  homologous  lines. 

EXERCISES. 

1.  If  the  area  of  a  rectangle  is  7956  sq.  in.,  and  its  base  3^  yd., 
find  its  perimeter  in  feet. 

2.  If  the  base  and  altitude  of  a  rectangle  are  14  ft.  7  in.,  and  5  ft. 
3  in.,  respectively,  what  is  the  side  of  an  equivalent  square  ? 

3.  Find  the  dimensions  of  a  rectangle  whose  area  is  168,  and 
perimeter  52. 

(Let  X  represent  the  base.) 


170  PLANE  GEOMETRY.— BOOK  IV. 

Prop.  VIII.    Theorem. 

321.  Two  triangles  having  an  angle  of  one  equal  to  an 
angle  of  the  other,  are  to  each  other  as  the  products  of  the 
sides  including  the  equal  angles. 


Given  Z  A  common  to  A  ABC  and  AB'C. 

To  Prove  ABC_^ABxAC^ 

AB'C     AB'xAC 

Proof.     Draw  line  B'C. 

Then  A  ABC  and  AB'C,  having  the  common  vertex  0, 
and  their  bases  AB  and  AB'  in  the  same  str.  line,  have  the 
same  altitude. 

And  A  AB'C  and  AB'C,  having  the  common  vertex  5, 
and  their  bases  AC  and  AC  in  the  same  str.  line,  have  the 
same  altitude. 

AB'C  ^  AC 
'  *  AB'C     AC 
Multiplying  these  equations,  we  have 

ABC      AB'C         ABC  ^  AB  x  AC 
AB'C     AB'C'    ^  AB'C     AB'  x  AC' 

EXERCISES. 

4.  The  area  of  a  rectangle  is  143  sq.  ft.  75  sq.  in.,  and  its  base  is 
3  times  its  altitude.    Find  each  of  its  dimensions. 

(Let  X  represent  the  altitude.) 

5.  The  hypotenuse  of  a  right  triangle  is  5  ft.  5  in.,  and  one  of  its 
legs  is  2  ft.  9  in.    Find  its  area. 


AREAS  OF  POLYGONS. 


171 


Prop.  IX.     Theorem. 

322.    Two  similar  polygons  are  to  each  other  as  the  squares 
of  their  homologous  sides. 


E4- 


Given  AB  and  A'B'  homologous  sides  of  similar  polygons 
AC  and  AC,  whose  areas  are  /fand  K\  respectively. 

K      AB" 


To  Prove 


K' 


A'B' 


Proof.    Draw  diagonals  EB,  EC,  E'B',  and  E'C. 
Then,  A  ABE  is  similar  to  A  A'B'E'. 

ABE        A& 


In  like  manner, 


AB'E'     jr^' 


(§  267) 
(§  319) 


and 


BCE 

BC^ 

AB" 

B'CE' 

B'C' 

A'B'' 

CDE 

CD' 

AB' 

C'D'E' 

CD'' 

A'B'' 

ABE 

BCE 

CDE 

A'B'E'     B'CE'     C'D'E' 
ABE  +  BCE  +  CDE  ABE 


A'B'E'  +  B'CE'  +  C'D'E'     A'B'E' 


(§  253,  2) 

(?) 

(§  240) 


K      ABE 


AB' 


K'     A'B'E'     A^> 


323.   Cor.     Two  similar  polygons  are  to  each  other  as  the 
squares  of  their  perimeters.  (§  268) 


172 


PLANE  GEOMETRY.  — BOOK  IV. 


Prop.  X.     Problem. 

324.    To  express  the  area  of  a  triangle  in  terms  of  its  three 
sides. 


Given  sides  BC,  CA,  and  AB,  of  A  ABC,  equal  to  a,  b, 
and  c,  respectively. 

Required  to  express  area  ABC  in  terms  of  a,  b,  and  c. 
Solution.     Let  C  be  an  acute  Z,  and  draw  altitude  AD. 

.:  c'  =  a'-\-b'-2axCD.    (§277) 
Transposing,       2a  x  CD  =  a^  -\' b^  —  c?. 

...  CD  =  ^^±^^^. 


2a 


(§  273) 


AU  =  AC'-CD" 

=  (AC -\- CD)  {AC -CD) 

^{2ab  +  a^  +  b^-  c")  (2 ab-a^-b''-\-  c") 

^  r(^  +  bY  -  c^l  [c^  -  (g  -  5)T 
4a2 

_  (g  +  6  +  c)  (g  +  ^  —  c)  (c  +  «  —  &)  (c  —  g  +  ^)      /-IN 

Now  let  a  +  6  +  c  =  2  s. 

^j^2^2g(2s-2c)(2s-26)(2s-2a) 

^16g(s-a)(g-6)(g-c) 
4a2 


AREAS  OF  POLYGONS.  173 


...  AI?  -  ^^^^^  ~  ^^  ^^  -b)(s-  c) 
a 
.',  area  ABC  =\axAD  (?) 

=  Vs(s  —  a){s  —  h)  (s  —  c). 

325.  Sch.    Let  it  be  required  to  find  the  area  of  a  tri- 
angle whose  sides  are  13, 14,  and  15. 

Let  a  =  13,  b  =  14,  and  c  =  15 ;  then 

s  =  i(13  +  14  +  15)  =  21. 

Whence,  s— a  =  8,  s  —  6  =  7,  and  s  —  c  =  6. 

Then,  the  area  of  the  triangle  is 


V21  x8x7x6  =  V3x7x23x7x2x3 

=  V2*  X  32  X  72  =  22  X  3  X  7  =  84. 


EXERCISES. 

6.  Find  the  area  of  a  triangle  whose  sides  are  8,  13,  and  15. 

7.  The  area  of  a  square  is  693  sq.  yd.  4  sq.  ft. ;  find  its  side. 

8.  If  the  altitude  of  a  trapezoid  is  1  ft.  4  in.,  and  its  bases  1  ft.  1  in. 
and  2  ft.  5  in.,  respectively,  what  is  its  area  ? 

9.  If,  in  figure  of  Prop.  VII.,  AB  =  9,  A'B'  =  7,  and  the  area  of 
A'B'C  is  147,  find  area  ABC. 

10.  If  the  sides  of  triangle  ABC  are  AB  =  25,  BC=  17,  and 
CA  =  28,  find  its  area,  and  the  length  of  the  perpendicular  from  each 
vertex  to  the  opposite  side. 

11.  Find  the  length  of  the  diagonal  of  a  rectangle  whose  area  is 
2640,  and  altitude  48. 

12.  Find  the  lower  base  of  a  trapezoid  whose  area  is  9408,  upper 
base  79,  and  altitude  96. 

13.  The  area  of  a  rhombus  is  equal  to  one-half  the  product  of  its 
diagonals.     (§117.) 

14.  The  diagonals  of  a  parallelogram  divide  it  into  four  equivalent 
triangles. 

15.  Lines  drawn  to  the  vertices  of  a  parallelogram  from  any  point 
in  one  of  its  diagonals  divide  the  figure  into  two  pairs  of  equivalent 
triangles.     (Ex.  63,  p.  67.) 

16.  The  area  of  a  certain  triangle  is  2|:  times  the  area  of  a  similar 
triangle.  If  the  altitude  of  the  first  triangle  is  4  ft.  3  in.,  what  is  the 
homologous  altitude  of  the  second  ?     (§  320.) 


174 


PLANE   GEOMETRY.— BOOK  IV. 


326.  Sch.  Since  the  area  of  a  square  is  equal  to  the 
square  of  its  side  (§  307),  we  may  state  Prop.  XXIV., 
Book  III.,  as  follows : 

In  any  right  triangle,  the  square  described  upon  the 
hypotenuse  is  equivalent  to  the  sum  of  the  squares  described 
upon  the  legs. 

The  theorem  in  the  above  form  may  be  proved  as  follows : 


y^  "i- 

\ 

; 

P 

B 


F  ME 

Given  ABEF,  AGGH,  and  BCKL  squares  described  upon 
hypotenuse  AB,  and  legs  AG  and  BG,  respectively,  of  rt. 
AABG. 
To  Prove  area  ABEF=  area  AGGH+  area  BGKL. 
Proof.     Draw  line  GD  ±  AB,  and  produce  it  to  meet  EF 
at  Jf ;  also,  draw  lines  BH  and  GF. 
Then  in  A  ABH  and  AGF,  by  hyp., 

AB  =  AF  and  AH=  AG. 
Also,  ZBAH  =  ZGAF, 

for  each  is  equal  to  a  rt.  Z  -\-  Z  BAG. 

.'.  A  ABH  =  A  AGF.  (?) 

Now  A  ABH  has  the  same  base  and  altitude  as  square 
AGGH. 

.-.  Sivesi  ABH =i3LTe3i  AGGH.  (§  315) 

And  A  AGF  has  the  same  base   and  altitude  as   rect. 
ADMF. 


AREAS  OF  POLYGONS. 


176 


But,  area  ABH  =  area  ACF. 

.'.  isiTesiACGH=i3iTea.ADMF,  (?) 

or  area  ^OG^ir=  area  ^/)Jfi^.  (1) 

Similarly,  by  drawing  lines  AL  and  CE,  we  may  prove 

area  BCKL  =  area  BDME.  (2) 

Adding  (1)  and  (2),  we  have 

area  ACGH  +  area  BCKL  =  area  ABEF. 

327.  Sch.  The  theorem  of  §  326  is  supposed  to  have 
been  first  given  by  Pythagoras,  and  is  called  after  him  the 
Pythagorean  Theorem. 

Several  other  propositions  of  Book  III.  may  be  put  in 
the  form  of  statements  in  regard  to  areas ;  as,  for  example. 
Props.  XXV.  and  XXVI. 


EXERCISES. 

17.  If  EF  is  any  straight  line  drawn  through 
the  centre  of  parallelogram  A  BCD,  meeting 
sides  AD  and  BC  at  E  and  F,  respectively, 
prove  triangles  BEF  and  CED  equivalent. 
(Ex.  61,  p.  6Q.)  A 

(Prove  BEDF  a  O  by  §  112.) 

18.  The  side  of  an  equilateral  triangle  is  5 ;  find  its  area. 
21,  p.  151.) 

19.  The  altitude  of  an  equilateral  triangle  is  3  ;  find  its  area, 

C 


(Ex. 


C 


20.  Two  triangles  are  equivalent  if  they 
have  two  sides  of  one  equal  respectively  to  two 
sides  of  the  other,  and  the  included  angles 
supplementary.  D' 

21.  One  diagonal  of  a  rhombus  is  five-thirds  of  the  other,  and  the 
difference  of  the  diagonals  is  8  ;  find  its  area.     (Ex.  13,  p.  173.) 

22.  If  D  and  E  are  the  middle  points  of  sides  BC  and  AC,  respec- 
tively, of  triangle  ABC,  prove  triangles  ABD  and  ABE  equivalent. 
(§  80.) 


176  PLANE   GEOMETRY.— BOOK  IV. 

23.  If  E  is  the  middle  point  of  CD,  one  of  the       j 
non-parallel  sides  of  trapezoid  ABCB,  and  a  par- 
allel to  AB  drawn  through  E  meets  EC  bX  F  and 
AD  at  G,  prove  parallelogram  ABFG  equivalent 
to  the  trapezoid.  ^ 

24.  The  sides  AB,  BC,  CD,  and  DA  of  quadrilateral  ABCD  are 
10,  17,  13,  and  20,  respectively,  and  the  diagonal  ^C  is  21.  Find  the 
area  of  the  quadrilateral. 

25.  Find  the  area  of  the  square  inscribed  in  a  circle  whose  radius 
is  3. 

(The  diagonal  is  a  diameter,  by  §  157.) 

26.  The  area  of  an  isosceles  right  triangle  is  81  sq.  in.;  find,  its 
hypotenuse  in  feet. 

(Represent  one  of  the  equal  sides  by  x.) 

27.  The  area  of  an  equilateral  triangle  is  9V3 ;  find  its  side. 
(Represent  the  side  by  x.) 

28.  The  area  of  an  equilateral  triangle  is  16\/3  ;  find  its  altitude. 
(Represent  the  altitude  by  x.) 

29.  The  base  of  an  isosceles  triangle  is  66,  and  each  of  the  equal 
sides  is  53  ;  find  its  area. 

/\ 

30.  The  area  of  a  triangle  is  equal  to  one-half  y>\-\ 
the  product  of  its  perimeter  by  the  radius  of  the         K^\     >. 
inscribed  circle.                                                               k  ^'1^, J\ 

^  D     ^^B 

31.  The  area  of  an  isosceles  right  triangle  is  equal  to  one-fourth 
the  area  of  the  square  described  upon  the  base.     (§  307.) 

32.  If  angle  A  of  triangle  ABC  is  30°,  prove 

area  ABC  =  ^  AB  x  AC, 
(Draw  CD  ±AB;  then  CD  may  be  found  by  Ex.  104,  p.  71.) 

33.  A  circle  whose  diameter  is  12  is  inscribed  in  a  quadrilateral 
whose  perimeter  is  50.     Find  the  area  of  the  quadrilateral. 

(Compare  Ex.  30,  p.  176.) 

34.  Two  similar  triangles  have  homologous  sides  equal  to  8  and  15, 
respectively.  Find  the  homologous  side  of  a  similar  triangle  equiva- 
lent to  their  sum.     (§  319.) 

35.  If  E  is  any  point  within  parallelogram  ABCD,  triangles  ABE 
and  CDE  are  together  equivalent  to  one-half  the  parallelogram. 

(Draw  through  ^  a  ||  to  AB.) 


AREAS  OF  POLYGONS. 


177 


36.  The  non-parallel  sides,  AB  and  CD,  of  a  trapezoid  are  each 
25  units  in  length,  and  the  sides  AD  and  BC  are  33  and  19  units, 
respectively.    Find  the  area  of  the  trapezoid. 

(Draw  through  ^  a  ||  to  CD,  and  a  ±  to  AD.) 

37.  If  the  area  of  a  polygon,  one  of  whose  sides  is  15  in.,  is  375 
sq.  in.,  what  is  the  area  of  a  similar  polygon  whose  homologous  side 
is  18  in.? 

38.  If  the  area  of  a  polygon,  one  of  whose  sides  is  36  ft.,  is  648 
sq.  ft.,  what  is  the  homologous  side  of  a  similar  polygon  whose  area 
is  392  sq.ft.? 


39.  If  one  diagonal  of  a  quadrilateral  bisects 
the  other,  it  divides  the  quadrilateral  into  two 
equivalent  triangles. 

(To  prove  A  ABC  =c=  A  ACD.) 

40.  Two  equivalent  triangles  have  a  com- 
mon base,  and  lie  on  opposite  sides  of  it.  Prove 
that  the  base,  produced  if  necessary,  bisects 
the  line  joining  their  vertices. 

(To  prove  CD  =  CD.) 


41.   If  the  sides  of  a  triangle  are  15,  41,  and  52,  find  the  radius  of 
the  inscribed  circle.     (Ex.  30,  p.  176.) 


find  its 


42.   The  area  of  a  rhombus  is  240,  and  its  side  is  17 
diagonals.     (Ex.  13,  p.  173.) 

(Represent  the  diagonals  by  2  x  and  2  y.) 


43.  The  sum  of  the  perpendiculars  from  any 
point  within  an  equilateral  triangle  to  the  three 
sides  is  equal  to  the  altitude  of  the  triangle. 

^  Te "C 

44.  The  longest  sides  of  two  similar  polygons  are  18  and  3,  respec- 
tively. How  many  polygons,  each  equal  to  the  second,  will  form  a 
polygon  equivalent  to  the  first  ?     (§  322.) 

45.  If  the  sides  of  a  triangle  are  25,  29,  and  36,  find  the  diameter 
of  the  circumscribed  circle.     (§  287.) 

(The  altitude  of  a  A  equals  its  area  divided  by  one-half  its  base.) 


178  PLANE   GEOMETRY.— BOOK  IV. 

46.  If  a  is  the  base,  and  6  one  of  the  equal  sides  of  an  isosceles 
triangle,  prove  its  area  equal  to  ^aVTpTT^i. 

47.  The  sides  AB  and  AC  of  triangle  ABC  are  15  and  22,  respec- 
tively. From  a  point  D  in  AB^  a  parallel  to  BC  ia  drawn  meeting 
AC  At  E,  and  dividing  the  triangle  into  tvsro  equivalent  parts.  Find 
AD  and  AE.     (§  319.) 

48.  The  segments  of  the  hypotenuse  of  a  right  triangle  made  by 
a  perpendicular  drawn  from  the  vertex  of  the  right  angle,  are  5|  and 
9|,  respectively  ;  find  the  area  of  the  triangle. 

49.  Any  straight  line  drawn  through  the 
centre  of  a  parallelogram,  terminating  in  a 
pair  of  opposite  sides,  divides  the  parallelo- 
gram into  two  equivalent  quadrilaterals. 
(Ex.  61,  p.  66.) 

50.  If  JS  is  the  middle  point  of  CD,  one  of  the  non-parallel  sides 
of  trapezoid  ABCD,  prove  triangle  ABE  equivalent  to  \ABCD. 

(Draw  through  JS"  a  ||  to  AB.) 

51.  The  sides  of  triangle  ABC  are  AB  =  IZ,  BC=U,  and 
CA  =  15.  If  AD  is  the  bisector  of  angle  A,  meeting  BC  at  D,  find 
the  areas  of  triangles  ABD  and  ACD.     (§§  249,  325.) 

52.  The  longest  diagonal  AD  of  pentagon  ABCDE  is  44,  and  the 
perpendiculars  to  it  from  B,  C,  and  E  are  24,  16,  and  15,  respectively. 
If  AB  =  25,  CD  =  20,  and  AE  =  17,  what  is  the  area  of  the  penta- 
gon ?     (§318.) 

53.  The  sides  of  a  triangle  are  proportional  to  the  numbers  7,  24, 
and  25,  respectively.  The  perpendicular  to  the  third  side  from  the 
vertex  of  the  opposite  angle  is  13 |i.     Find  the  area  of  the  triangle. 

(Represent  the  sides  by  7  x,  24  x,  and  25  x,  respectively  ;  the  A 
is  a  rt.  A  by  Ex.  63,  p.  154.) 

54.  If  E  and  F  are  the  middle  points  of  sides  AB  and  AC,  respec- 
tively, of  a  triangle,  and  D  is  any  point  in  BC,  prove  quadrilateral 
AEDF  equivalent  to  one-half  triangle  ABC. 

(Vro\e  A  DEF  =0=1  A  ABC,  by  aid  of  Ex.  64,  p.  67.) 

55.  If  E,  F,  Cr,  and  H  are  the  middle  points 
of  sides  AB,  BC,  CD,  and  DA,  respectively,  of 
quadrilateral  ABCD,  prove  EFCH  a  parallelo- 
gram equivalent  to  one-half  ABCD. 

(By  Ex.  64,  p.  67,  area  EBF  -  \  area  ABC.)  ^ 

Note.     For  additional  exercises  on  Book  IV. ,  see  p,  229. 


AREAS  OF  POLYGONS. 


179 


CONSTRUCTIONS. 

Prop.  XI.     Problem. 

328.    To  construct  a  square  equivalent  to  the  sum  of  two 
given  squares. 


M 

c 

N 

A  B 

Given  squares  M  and  N. 

Required  to  construct  a  square  =c=M-\-N. 

Construction.     Draw  line  AB  equal  to  a  side  of  M. 

At  A  draw  line  ACA.AB,  and  equal  to  a  side  of  N\  and 
draw  line  EG. 

Then,  square  P,  described  with  its  side  equal  to  EC,  will 
be  =0=  Jf+JV. 

Proof.     In  rt.  A  ABC,  BG^  =  Jff  +  AC"^-  (?) 

.-.  area  P  —  area  M-\-  area  N.      (§  307) 

329.  Cor.  By  an  extension  of  the  above  method,  a  square 
may  be  constructed  equivalent  to  the  sum  of  any  number  of 
given  squares. 

Given  three  squares  whose  sides  are  equal  to  ^ 

m,  n,  and  p,  respectively. 

Required  to  construct  a  square  =o=  the  sum  of 
the  given  squares. 

Construction.     Draw  line  AB  =  m. 

Draw  line  AC  1.  AB,  and  equal  to  n,  and 
line^O.  ^^     '"     ^ 

Draw  line  CD  ±  EC,  and  equal  to  p,  and  line  ED. 

Then,  the  square  described  with  its  side  equal  to  ED  will 
be  ^  the  sum  of  the  given  squares. 

(The  proof  is  left  to  the  pupil.) 


P/ 


CL 


180 


PLANE   GEOMETRY.— BOOK  IV. 


Prop.  XII.     Problem. 

330.    To  construct  a  square  equivalent  to  the  difference  of 
two  given  squares. 


N 


B 


Given  squares  M  and  N,  M  being  >  N. 

Required  to  construct  a  square  =o-  M—  N. 

Proof.     Draw  the  indefinite  line  AD. 

At  A  draw  line  AB  _L  AD,  and  equal  to  a  side  of  W. 

With  5  as  a  centre,  and  with  a  radius  equal  to  a  side  of 
M,  describe  an  aj-c  cutting  AD  at  C,  and  draw  line  BC. 

Then,  square  P,  described  with  its  side  equal  to  AC,  will 
be  ^M-N. 

Proof.     In  rt.  A  ABC,  AC''  =  BG^  -  Aff.  (?) 

.-.  area  F=  area  M—  area  N.  (?) 


Prop.  XIII.     Problem. 

331.    To  construct  a  square  equivalent  to  a  given  paral- 
lelogram. 

K If 

C  ' 


A     E 


B 


F  G 

Given  CJABCD. 

Required  to  construct  a  square  =c=  ABCD. 

Construction.  Draw  line  DE  ±  AB,  and  construct  line 
FG  a  mean  proportional  between  lines  AB  and  DE  (§  292). 

Then,  square  FGHK,  described  with  its  side  equal  to  FG, 
will  be  ^  ABCD. 


AREAS  OF  POLYGONS.  181 

Proof.     By  cons.,     AB:  FG  =  FG:  DE. 

.-.  FG^  =  ABxDE.  (?) 

.'.  Sivesi  FGHK  =  sivesi  ABCD.  (?) 

332.  Cor.  A  square  may  be  constructed  equivalent  to  a 
given  triangle  by  taking  for  its  side  a  mean  proportional 
between  the  base  and  one-half  the  altitude  of  the  triangle. 


Ex.  56.  To  construct  a  triangle  equivalent  to  a  given  square, 
having  given  its  base  and  an  angle  adjacent  to  the  base. 

(Take  for  the  required  altitude  a  third  proportional  to  one-half  the 
given  base  and  the  side  of  the  given  square.) 

Prop.  XIV.    Problem. 

333.  To  construct  a  rectangle  equivalent  to  a  given  square^ 
having  the  sum  of  its  base  and  altitude  equal  to  a  given  line. 


M 


X>- 


N 


Given  square  M,  and  line  AB. 

Required  to  construct  a  rectangle  =c=  M,  having  the  sum 
of  its  base  and  altitude  equal  to  AB. 

Construction.  With  AB  as  a  diameter,  describe  semi- 
circumference  ADB. 

Draw  line  AC  _L  AB,  and  equal  to  a  side  of  M. 

Draw  line  CF  II  AB,  intersecting  arc  ADB  at  D,  and 
line  DE  A.  AB. 

Then,  rectangle  N,  constructed  with  its  base  and  altitude 
equal  to  BE  and  AE,  respectively,  will  be  =c=  M. 

Proof.  AE:DE=DE:  BE.  (§  270,  1) 

.-.  AE  xBE  =  DE'  =  AC\  (?) 

.-.  area  N=  area  M.  (?) 


182  PLANE   GEOMETRY.— BOOK  IV, 


Prop.  XV.     Problem. 

334.  To  construct  a  rectangle  equivalent  to  a  given  square, 
having  the  difference  of  its  base  and  altitude  equal  to  a  given 
line. 


\    )p. ..^ 


N 


- E 

Given  square  M,  and  line  AB. 

Required  to  construct  a  rectangle  =0=  M,  having  the  differ- 
ence of  its  base  and  altitude  equal  to  AB. 

Construction.     With  AB  as  a  diameter,  describe  O  ADB. 

Draw  line  AC  J-  AB,  and  equal  to  a  side  of  M. 

Through  centre  0  draw  line  CO,  intersecting  the  circum- 
ference at  D  and  E. 

Then,  rectangle  N,  constructed  with  its  base  and  altitude 
equal  to  CE  and  CD,  respectively,  will  be  =0=  M. 

Proof.  CE-CD  =  DE  =  AB.  (?) 

That  is,  the  difference  of  the  base  and  altitude  of  JV  is 
equal  to  AB. 

Again,  AC  is  tangent  to  O  ADB  at  A.  (?) 

.-.  CDxCE=  GA".  (§  282) 

.-.  area  JV=  area  M.  (?) 

EXERCISES. 

57.  To  construct  a  triangle  equivalent  to  a  given  triangle,  having 
given  its  base. 

(Take  for  the  required  altitude  a  fourth  proportional  to. the  given 
base,  and  the  base  and  altitude  of  the  given  A.) 
How  many  different  A  can  be  constructed  ? 

58.  To  construct  a  rectangle  equivalent  to  a  given  rectangle,  hav- 
ing given  its  base. 


AREAS  OF  POLYGONS. 


183 


59.   To  construct  a  square  equivalent  to  twice  a  given  square. 
(§  307.) 

Prop.  XVI.    Problem. 

335.    To  construct  a  square  having  a  given  ratio  to  a  given 
square. 

C. 


M 


m 


/  i  N 


\ 


\A 

N 

\^\ 

n       B 

!/         I 
A"~m"'l)" 


Given  square  M,  and  lines  m  and  n. 

Required  to  construct  a  square  having  to  M  the  ratio 
n:m. 

Construction.     On  line  AB,  take  AD  =  m  and  DB  —  n. 

With  AB  as  a  diameter,  describe  semi-circumference 
ACB. 

Draw  line  DC1.AB,  meeting  arc  ACB  at  G,  and  lines 
AC  and  BC 

0\i  AC  take  CE  equal  to  a  side  of  M;  and  draw  line 
EF II  AB,  meeting  BC  at  F. 

Then,  square  N,  constructed  with  its  side  equal  to  GF, 
will  have  to  M  the  ratio  n-.m. 

Proof.     Z  ACB  is  a  rt.  Z.  (?) 

Then  since  CD  is  ±  AB, 

AG^     AB  X  AD     AD     m  /^  271  2) 

(?) 


BC'     ABxBD     BD 

n 

since  EF  is  ||  AB, 

GE     AG 
GF     BC 

ge'   aW 

_m 

cf'    bg' 

n 

area  M     m 

area  iV      n 


(?) 


184  PLANE  GEOMETRY.— BOOK  IV. 

Prop.  XVII.    Problem. 

336.   To  construct  a  triangle  equivalent  to  a  given  polygon. 
A 


O       D 
Given  polygon  ABGDE. 
Required  to  construct  a  A  =c=  ABODE. 

Construction.  Take  any  three  consecutive  vertices,  as  Ay 
B,  and  C,  and  draw  diagonal  AC;  also,  line  BF  )|  AC,  meet- 
ing DC  produced  at  F,  and  line  AF. 

Then,  AFDE  is  a  polygon  =c=  ABODE,  having  a  number 
of  sides  less  by  one. 

Again,  draw  diagonal  AD ;  also,  line  EG  ||  AD,  meeting 
CD  produced  at  G,  and  line  AG. 

Then,  AFG  is  a  A  =0=  ABODE. 

Proof.     A  ABC  and  AFC  have  the  same  base  AC. 
And  since  their  vertices  B  and  F  lie  in  the  same  line  ||  to 
AC,  they  have  the  same  altitude.  (§  80) 

.-.  area  JjB(7  =  area  ^F(7.  (?) 

Adding  area  AODE  to  both  members,  we  have 

area  ABODE  =  area  AFDE. 

Again,  A  AED  and  AGD  have  the  same  base  AD,  and 
the  same  altitude. 

.-.  Sivesi  AED  =  Sivesi  AGD.  (?) 

Adding  area  AFD  to  both  members,  we  have 

area  AFDE  =  area  AFG. 

.-.  area  ABODE  =  area  AFG.  (?) 


AREAS   OF  POLYGONS. 


185 


Note.     By  aid  of  §§  336  and  332,  a  square  may  be  constructed 
equivalent  to  a  given  polygon. 


Prop.  XVIII.     Problem. 

337.    To  construct  a  polygon  similar  to  a  given  polygon,  and 
having  a  given  ratio  to  it. 


m 


Given  polygon  AC,  and  lines  m  and  n. 

Required  to  construct  a  polygon  similar  to  AC,  and  hav- 
ing to  it  the  ratio  n  :  m. 

Construction.  Construct  A'B^  the  side  of  a  square  having 
to  the  square  described  upon  AB  the  ratio  n :  m.         (§  335) 

Upon  side  A!B\  homologous  to  AB,  construct  polygon 


A^O  similar  to  polygon  AC. 

Then,  A^O  will  have  to  ^C  the  ratio  n  :  m. 
Proof.     Since  AC  is  similar  to  A'C, 
AC       Aff 


But  by  cons., 


A'C 

A'B'' 

AB' 

_m^ 

A'B'' 

n 

AC 

A'C 

n 

(§  295) 


(§  322) 


(?) 


Ex.  60.  To  construct  an  isosceles  triangle 
equivalent  to  a  given  triangle,  having  its  base  co- 
incident with  a  side  of  the  given  triangle. 


186  PLANE   GEOMETRY. —  BOOK  IV. 


Prop.  XIX.     Problem. 

338.    To  construct  a  polygon  similar  to  one  of  two  given 
polygons^  and  equivalent  to  the  other. 


Given  polygons  M  and  N. 

Required  to  construct  a  polygon  similar  to  M,  and  =0  N. 

Construction.     Let  AB  be  any  side  of  M. 

Construct  m,  the  side  of  a  square  =0=  M,  and  n,  the  side  of 
a  square  =0=  N.  (Note,  p.  185) 

Construct  A'B\  a  fourth  proportional  to  m,  n,  and  AB. 

Upon  side  A'B\  homologous  to  AB,  construct  polygon  P 
similar  to  M.  (§  295) 

Then,  P  =c=  JV. 

Proof.     Since  M  is  similar  to  P, 


area  M     AB 


(?) 


(?) 


area  P      A'B' 

But  by  cons.,  m  :  n  =  AB :  A'B',  or  ——- :  =  —  • 
^  '  '       A'B'      n 

area M _'m^_  area M 

area  P      n^      area  If 

.:  area  P  =  area  M 

EXERCISES. 

61.  To  construct  a  triangle  equivalent  to  a  given  square,  having 
given  its  base  and  the  median  drawn  from  the  vertex  to  the  base. 

(Draw  a  |1  to  the  base  at  a  distance  equal  to  the  altitude  of  the  A. ) 
What  restriction  is  there  on  the  values  of  the  given  lines  ? 

62.  To  construct  a  rhombus  equivalent  to  a  given  parallelogram, 
having  one  of  its  diagonals  coincident  with  a  diagonal  of  the  paral- 
lelogram.    (Ex.  60.) 


AREAS  OP  POLYGONS.  Jg*^ 

63.  To  draw  through  a  given  point  within  a  parallelogram  a  straight 
line  dividing  it  into  two  equivalent  parts.     (Ex,  49,  p,  178.) 

64.  To  construct  a  parallelogram  equivalent  to  a  given  trapezoid, 
having  a  side  and  two  adjacent  angles  coincident  with  one  of  the  non- 
parallel  sides  and  the  adjacent  angles,  respectively,  of  the  trapezoid 
(Ex.  23,  p.  176.) 

65.  To  construct  a  triangle  equivalent  to  a  given  triangle,  having 
given  two  of  its  sides.     (Ex.  67.) 

(Let  m  and  n  be  the  given  sides,  and  take  m  as  the  base.) 
Discuss  the  solution  when  the  altitude  is  <  n.     =  n.     >  w. 

66.  To  construct  a  right  triangle   equivalent  to  a  given  square, 
having  given  its  hypotenuse.     (Ex.  96,  p.  119.) 

(Find  the  altitude  as  in  Ex.  56. ) 

What  restriction  is  there  on  the  values  of  the  given  parts  ? 

67.  To  construct  a  right  triangle  equivalent  to  a  given  triangle, 
having  given  its  hypotenuse. 

What  restriction  is  there  on  the  values  of  the  given  parts  ? 

68.  To  construct  an  isosceles  triangle  equivalent  to  a  given  tri- 
angle, having  given  one  of  its  equal  sides  equal  to  m. 

(Draw  a  II  to  the  given  side  at  a  distance  equal  to  the  altitude.) 
Discuss  the  solution  when  the  altitude  is  <  m.     =  m.     >  m. 

69.  To  draw  a  line  parallel  to  the  base  of  a  /\^ 
triangle  dividing  it  into  two  equivalent  parts.             /   \. 

(§  319.)  b/ Xc 

(^AABC  and  AB'C  are  similar.)  g/ \^ 

70.  To  draw  through  a  given  point  in  a  side  of  a  parallelogram  a 
straight  line  dividing  it  into  two  equivalent  parts. 

71.  To  draw  a  straight  line  perpendicular  to  the  bases  of  a  trape- 
zoid, dividing  the  trapezoid  into  two  equivalent  parts. 

(A  str.  line  connecting  the  middle  points  of  the  bases  divides  the 
trapezoid  into  two  equivalent  parts.) 

72.  To  draw  through  a  given  point  in  one  of  the  bases  of  a  trape- 
zoid a  straight  line  dividing  the  trapezoid  into  two  equivalent  parts. 

(A  str.  line  connecting  the  middle  points  of  the  bases  divides  the 
trapezoid  into  two  equivalent  parts.) 

73.  To  construct  a  triangle  Similar  to  two  given  similar  triangles, 
and  equivalent  to  their  sum. 

(Construct  squares  equivalent  to  the  A.) 

74.  To  construct  a  triangle  similar  to  two  given  similar  triangles, 
and  equivalent  to  their  difference. 


Book  Y. 


REGULAR  POLYGONS. -MEASUREMENT  OP 
THE  CIRCLE. 

339.  Def.     A  regular  polygon  is  a  polygon  which  is  both 
equilateral  and  equiangular. 

Prop.  I.     Theorem. 

340.  A  circle  can  he  circumscribed  about,  or  inscribed  in, 
any  regular  polygon. 


Given  regular  polygon  ABODE. 

To  Prove  that  a  O  can  be  circumscribed  about,  or  inscribed 
in,  ABODE. 

Proof.  Let  0  be  the  centre  of  the  circumference  described 
through  vertices  A,  B,  and  0  (§  223). 

Draw  radii  OA,  OB,  00,  and  OD. 

In  A  OAB  and  OOD,     OB  =  00.  (?) 

And  since,  by  def.,  polygon  ABODE  is  equilateral, 

AB=OD. 

188 


REGULAR  POLYGONS.  j^og 

Again,  since,  by  def.,  polygon  ABODE  is  equiangular, 

Z.ABC=Z.BGD. 
And  since  A  OBC  is  isosceles, 

ZOBC=ZOCB.  (?) 

.-.  Z  ABC  -  Z  OBC=Z  BCD  -  Z  OOB. 
Or,  Z  Oi^^  =  Z  OCD. 

.-.  AOAB  =  AOCD.  (?) 

.-.  0^=0Z>.  (?) 

Then,  the  circumference  which  passes  through  A,  B,  and 
C  also  passes  through  D. 

In  like  manner,  it  may  be  proved  that  the  circumference 
which  passes  through  B,  C,  and  D  also  passes  through  E. 

Hence,  a  O  can  be  circumscribed  about  ABCDE. 

Again,  since  AB,  BC,  CD,  etc.,  are  equal  chords  of  the 
circumscribed  O,  they  are  equally  distant  from  0.      (§  164) 

Hence,  a  O  described  with  0  as  a  centre,  and  a  line  OF 
±  to  any  side  AB  as  a  radius,  will  be  inscribed  in  ABCDE. 

341.  Def.  The  centre  of  a  regular  polygon  is  the  common 
centre  of  the  circumscribed  and  inscribed  circles. 

The  angle  at  the  centre  is  the  angle  between  the  radii 
drawn  to  the  extremities  of  any  side  ;  as  AOB. 

The  radius  is  the  radius  of  the  circumscribed  circle,  OA. 
The  apothem  is  the  radius  of  the  inscribed  circle,  OF. 

342.  Cor.   From  the  equal  A  OAB,  OBC,  etc.,  we  have 

ZAOB  =  Z  BOO  =  Z  COD,  etc.  (?) 

But  the  sum  of  these  A  is  four  rt.  A.  (§  35) 

Whence,  the  angle  at  the  centre  of  a  regular  polygon  is  equal 
to  four  right  angles  divided  by  the  number  of  sides. 

EXERCISES. 
Find  the  angle,  and  the  angle  at  the  centre, 
1.    Of  a  regular  pentagon. 


190  PLANE   GEOMETRY.  — BOOK   V. 

2.  Of  a  regular  dodecagon. 

3.  Of  a  regular  polygon  of  32  sides. 

4.  Of  a  regular  polygon  of  25  sides. 

Prop.  II.     Theorem. 

343.   If  the  circumference  of  a  circle  he  divided  into  any 
number  of  equal  arcs, 

I.    Their  chords  form  a  regular  inscribed  polygon. 
II.    Tangents  at  the  points  of  division  form  a  regular  cir- 
cumscribed polygon. 

L A        F 


Given  circumference  ACD  divided  into  five  equal  arcs, 
AB,  BC,  CD,  etc.,  and  chords  AB,  BC,  etc. 

Also,  lines  LF,  FG,  etc.,  tangent  to  O  ACD  at  A,  B,  etc., 
respectively,  forming  polygon  FGHKL. 

To  Prove  polygons  ABCDE  and  FGHKL  regular. 

Proof.     Chord  AB  =  chord  BC  =  chord  CD,  etc.     (§  158) 

Again,       arc  BCDE  =  arc  CDEA  =  arc  DEAB,  etc., 
for  each  is  the  sum  of  three  of  the  equal  arcs  AB,  BC,  etc. 
.-.  Z  EAB  =  Z  ABC  =  Z  BCD,  etc.  (§  193) 

Therefore,  polygon  ABCDE  is  regular.  (§  339) 

Again,  in  A  ABF,  BCG,  CDH,  etc.,  we  have 
AB  =  BO=CD,QtG. 

Also,  since  arc  AB  =  arc  BC=  arc  CD,  etc.,  we  have 

Z  BAF=ZABF=  Z  CBG  =  Z  BCG,  etc.    (§  197) 

Whence,  ABF,  BCG,  etc.,  are  equal  isosceles  A.  (§§  68, 96) 


REGULAR  POLYGONS.  j^gj 

and  BF=  BG=CG=  OH,  etc.  (§  66) 

.-.  FG=GH=HK,  etc. 

Therefore,  polygon  FGHKL  is  regular.  (?) 

344.  Cor.  I.  1.  If  from  the  middle  point  of  each  arc  sub- 
tended by  a  side  of  a  regular  inscribed  polygon  lines  be  drawn 
to  its  extremities,  a  regular  inscribed  polygon  of  double  the 
number  of  sides  is  formed. 

2.  If  at  the  middle  2?oint  of  each  arc  included  between  two 
consecutive  points  of  contact  of  a  regular  circwnscribed  poly- 
gon tangents  be  drawn,  a  regular  circumscribed  polygon  of 
double  the  number  of  sides  is  formed. 

345.  Cor.  II.  An  equilateral  2'>olygon' inscribed  in  a  circle 
is  regular;  for  its  sides  subtend  equal  arcs.  (?) 

Prop.  III.     Theorem. 

346.  Tangents  to  a  circle  at  the  middle  points  of  the  arcs 
subtended  by  the  sides  of  a  regular  inscribed  polygon,  form 
a  regular  circumscribed  polygon. 

A' 


Given  ABODE  a  regular  polygon  inscribed  in  OAG,  and 
A'B'O'D'E'  a  polygon  whose  sides  A'B',  B'C',  etc.,  are 
tangent  to  O  AC  at  the  middle  points  F,  G,  etc.,  of  arcs 
AB,  BO,  etc.,  respectively. 

To  Prove  A'B'O'D'E'  a  regular  polygon. 

(Arc  AF=  arc  BF=  arc  BG  =  arc  OG,  etc.,  and  the  propo- 
sition follows  by  §  343,  II.) 


192 


PLANE  GEOMETRY.  — BOOK  V. 


Prop.  IV.     Theorem. 

347.   Regular  polygons  of  the  same  number  of  sides  are 
similar. 

D 

D' 


(The  polygons  fulfil  the  conditions  of  similarity  given  in 
§  252.) 

Prop.  V.     Theorem. 

348.  The  perimeters  of  two  regular  polygons  of  the  same 
number  of  sides  are  to  each  other  as  their  radii,  or  as  their 
apothems. 


F'    B' 


Given  P  and  P'  the  perimeters,  R  and  R'  the  radii,  and 
r  and  r'  the  apothems,  respectively,  of  regular  polygons  AC 
and  A^O  of  the  same  number  of  sides. 


To  Prove 


P 
P' 


R 
R' 


Proof.     Let  0  be  the  centre  of  polygon  AC,  and  0'  of 
A'C,  and  draw  lines  OA,  OB,  O'A',  and  O'B'. 
Also,  draw  line  OF  ±  AB,  and  line  O'F'  _L  A'B'. 
Then,  OA  =  R,  OA'  =  R',  OF  =  r,  and  O'P'  =  r'. 
Now  in  isosceles  A  OAB  and  OA'B', 

ZAO£  =  ZA'0'B',  (§342) 


BEGULAE  POLYGONS. 


193 


And  since  OA  =  OB  and  O'A'  =  O'B',  we  have 

OA  ^  OB 

O'A'      O'B'' 
Therefore,  A  OAB  and  O'A'B'  are  similar.  (§  261) 

^^      ^      "*  (§§253,11,264) 


A'B'     R'     r' 
But  polygons  AC  and  A'O  are  similar. 
P  ^  AB 
' '  P'     A'B'' 

"  P'     R'     r'' 


(§  347) 
(§  268) 

(?) 


349.  Cor.     Let  ^denote  the  area  of  polygenic,  and  J^T' 
of  A'C. 

^^      ^  (§  322) 


But, 


A'B'      R 


K'     i2'2 


That  is,  the  areas  of  two  regular  polygons  of  the  same 
number  of  sides  are  to  each  other  as  the  squares  of  their 
radii,  or  as  the  squares  of  their  apothems. 


Prop.  VI.     Theorem. 

350.    The  area,  of  a  regular  polygon  is  equal  to  one-half 
the  product  of  its  perimeter  and  apothem. 

D 


A       F       B 
Given  the  perimeter  equal  to  P,  and  the  apothem   OF 
equal  to  r,  of  regular  polygon  AC. 
To  Prove  area  AC  =  ^Pxr. 

(A  OAB,  OBC,  etc.,  have  the  common  altitude  r.) 


194 


PLANE   GEOMETRY.  — BOOK  V. 


Prop.  VII.     Problem. 
351.    To  inscribe  a  square  m  a  given  circle. 

B 


Given  O  AG. 
Required  to  inscribe  a  square  in  O  AC. 
Construction.     Draw  diameters  AC  and  BD  ±  to  each 
other,  and  chords  AB,  BC,  CD,  and  DA. 
Then,  ABCD  is  an  inscribed  square. 
(The  proof  is  left  to  the  pupil ;  see  §  343,  I.) 


352.   Cor.   Denoting  radius  OA  by  R,  we  have 

AB'  =  6A^  +  OB''  =^2  R\  (§  272) 

.-.  AB  =  BV2. 

That  is,  the  side  of  an  inscribed  square  is  equal  to  the 
radius  of  the  circle  multiplied  by  V2. 


Prop.  VIII.     Problem. 

353.    To  inscribe  a  regular  hexagon  in  a  given  circle. 
B^ .    C 


Given  O  AG. 


REGULAR  POLYGONS.  I95 

Required  to  inscribe  a  regular  hexagon  in  O  ^0. 

Construction.     Draw  any  radius  OA. 

With  ^  as  a  centre,  and  AG  as  a  radius,  describe  an  arc 
cutting  the  given  circumference  at  B,  and  draw  chord  AB. 

Then,  AB  is  a  side  of  a  regular  inscribed  hexagon. 

Hence,  to  inscribe  a  regular  hexagon  in  a  given  O,  apply 
the  radius  six  times  as  a  chord. 

Proof.     Draw  radius  OB ;  then,  A  OAB  is  equilateral.  (?) 

Therefore,  A  OAB  is  equiangular.  (§  95) 

Whence,  Z  AOB  is  one-third  of  two  rt.  A.  (?) 

Then,  Z  AOB  is  one-sixth  of  four  rt.  A,  and  arc  AB  is 
one-sixth  of  the  circumference.  (§  154) 

Then,  AB  is  a  side  of  a  regular  inscribed  hexagon. 

(§  343,  I.) 

354.  Cor.  I.  The  side  of  a  regular  inscribed  hexagon  is 
equal  to  the  radius  of  the  circle. 

355.  Cor.  II.  If  chords  be  drawn  joining  the  alternate 
vertices  of  a  regular  inscribed  hexagon,  there  is  formed  an 
inscribed  equilateral  triangle. 

356.  Cor.  ni.     The    side   of  an   in-  ^ — ^^ 
scribed     equilateral    triangle     is     equal         /  ^.^  \\ 

to   the   radius    of  the    circle   multiplied     .  L^__ .^  (j 

by   V3.  r  J 

Given  AB  a  side  of  an  equilateral  A  y^    ^x 

inscribed  in  O  AD  whose  radius  is  B.  j) 

To  Prove  AB  =  R  V3. 

Proof.     Draw  diameter  AO,  and  chord  BC,  then,  BC  is 

a  side  of  a  regular  inscribed  hexagon.  (§  355) 

Now  ABC  is  art.  A.  (§  195) 

.:  AB'=AC'-BC'  (?) 

=={2Ey-R'  (§354) 

=  AR'-Ii^  =  3Ii'. 

.:  ulB  =  i2V3. 


196 


PLANE   GEOMETRY.  — BOOK  V. 


Prop.  IX.     Problem. 
357.    To  inscribe  a  regular  decagon  in  a  given  circle. 


Given  OAG. 
Required  to  inscribe  a  regular  decagon  in  O  AC. 
Construction.     Draw  any  radius  OA.  and  divide  it  inter- 
nally in  extreme  and  mean  ratio  at  ilf  (§  297),  so  that 

OA'.OM=OM:AM.  (1) 

With  ^  as  a  centre,  and  OM  as  a  radius,  describe  an  arc 
cutting  the  given  circumference  at  B,  and  draw  chord  AB. 
Then,  AB  is  a  side  of  a  regular  inscribed  decagon. 
Hence,  to  inscribe  a  regular  decagon  in  a  given  O,  divide 
the  radius  internally  in  extreme  and  mean  ratio,  and  apply 
the  greater  segment  ten  times  as  a  chord. 
Proof.     Draw  lines  OB  and  BM. 
In  A  OAB  and  ABM,  ZA  =  ZA. 
And  since,  by  cons.,  0M=  AB,  the  proportion  (1)  becomes 

OA:AB  =  AB:AM. 
Therefore,  A  OAB  and  ABM  are  similar. 

.-.  ZABM=ZAOB. 
Again,  A  OAB  is  isosceles. 
Hence,  the  similar  AABMh  isosceles,  and 
AB  =  BM=OM, 
.'.  ZOBM=ZAOB. 
.-.  Z  ABM-h  Z  OBM=  Z  AOB  -f-  Z  AOB. 


(§261) 
(?) 
(?) 

(Ax.  1) 
(?) 


REGULAR  POLYGONS.  I97 

Or,  Z0BA  =  2ZA0B.  (2) 

But  since  A  OAB  is  isosceles, 

2  Z  OBA  +  Z  AOB  =  180°.  (§  84) 

Then,  by  (2),        5  ZAOB  =  180°,  or  Z  AOB  =  36°. 
Therefore,  ZAOB  is  one-tenth  of  four  rt.  A,  and  AB  is  a 
side  of  a  regular  inscribed  decagon.  (?) 

358.  Cor.  I.  If  chords  be  drawn  joining  the  alternate  ver- 
tices of  a  regular  inscribed  decagon,  there  is  formed  a  regular 
inscribed  pentagon. 

359.  Cor.  II.  Denoting  the  radius  of  the  O  by  R,  we 
have 

AB  =  0M=  -^(V^-1).  (§  298) 

This  is  an  expression  for  the  side  of  a  regular  inscribed 
decagon  in  terms  of  the  radius  of  the  circle. 

Prop.  X.     Problem. 

360.  To  construct  the  side  of  a  regular  pentedecagon  in- 
scribed in  a  given  circle. 


Given  arc  MN. 

Required  to  construct  the  side  of  a  regular  inscribed 
polygon  of  fifteen  sides. 

Construction.  Construct  chord  AB  a  side  of  a  regular 
inscribed  hexagon  (§  353),  and  chord  AC  a  side  of  a  regular 
inscribed  decagon  (§  357),  and  draw  chord  BC. 

Then,  BC  is  a  side  of  a  regular  inscribed  pentedecagon. 

Proof.  By  cons.,  arc  BC  is  l  —  ^,  or  -^,  of  the  circum- 
ference. 

Hence,  chord  BO  is  a  side  of  a  regular  inscribed  pente- 
decagon. (?) 


198  PLANE   GEOMETRY.— BOOK  V. 

361.  Sch.  I.  By  bisecting  arcs  AB,  BC,  etc.,  in  the  figure 
of  Prop.  VII.,  we  may  xjonstruct  a  regular  inscribed  octagon 
(§  343,  I.);  and  by  continuing  the  bisection,  we  may  con- 
struct regular  inscribed  polygons  of  16,  32,  64,  etc.,  sides. 

In  like  manner,  by  aid  of  Props.  VIII.,  IX.,  and  X.,  we 
may  construct  regular  inscribed  polygons  of  12,  24,  48,  etc., 
or  of  20,  40,  80,  etc.,  or  of  30,  60,  120,  etc.,  sides. 

362.  Sch.  II.  By  drawing  tangents  to  the  circumference 
at  the  vertices  of  any  one  of  the  above  inscribed  regular 
polygons,  we  may  construct  a  regular  circumscribed  polygon 
of  the  same  number  of  sides.  (§  343,  II.) 

EXERCISES. 

5.  The  angle  at  the  centre  of  a  regular  polygon  is  the  supplement 
of  the  angle  of  the  polygon.     (§  127.) 

6.  The  circumference  of  a  circle  is  greater  than  the  perimeter  of 
any  inscribed  polygon. 

7.  An  equiangular  polygon  circumscribed  about  a  circle  is  regular. 
(§  202.) 

If  r  represents  the  radius,  a  the  apothem,  s  the  side,  and  k  the  area, 

8.  In  an  equilateral  triangle,  a  =  ^r,  and  k  =  fr^v'S. 

9.  In  a  square,  a  =  ^  r\/2,  and  k  =  2  r^. 

10.  In  a  regular  hexagon,  a  =  ^rVS,  and  k  =  fj-^VS. 

11.  In  an  equilateral  triangle,  r  =  2  a,  s  =  2a  v'3,  and  k  =  3 a^VS. 

12.  In  a  square,  r  =  a \/2,  s  =  2a,  and  k  =  ia"^. 

13.  In  a  regular  hexagon,  r  =  |  aa/S,  and  k  =  2  a^VS. 

14.  In  an  equilateral  triangle,  express  r,  a,  and  k  in  terms  of  s. 

15.  In  a  square,  express  r,  a,  and  k  in  terms  of  s. 

16.  In  a  regular  hexagon,  express  a  and  k  in  terms  of  s. 

17.  In  an  equilateral  triangle,  express  r,  a,  and  s  in  terms  of  k. 

18.  In  a  square,  express  r,  a,  and  s  in  terms  of  k. 

19.  In  a  regular  hexagon,  express  r  and  a  in  terms  of  k. 

20.  The  apothem  of  aij  equilateral  triangle  is  one-third  the  altitude 
of  the  triangle, 


REGULAR  POLYGONS. 


199 


21.  The  sides  of  a  regular  polygon  circumscribed  about  a  circle 
are  bisected  at  the  points  of  contact.     (§  94.) 

22.  The  radius  drawn  from  the  centre  of  a  regular  polygon  to  any 
vertex  bisects  the  angle  at  that  vertex.     (§  44.) 

23.  The  diagonals  of  a  regular  pentagon  are  equal.     (§03.) 

B 

24.  The  figure  bounded  by  the  five  diagonals 
of  a  regular  pentagon  is  a  regular  pentagon.  h^ — ^ — X^ — ^c 

(Prove,  by  aid  of  §  164,  that  a  O  can  be  in- 
scribed in  FGHKL  ;  then  use  Ex.  7,  p.  198.) 

25.  The  area  of  a  regular  inscribed  hexagon  is  a  mean  propor- 
tional between  the  areas  of  an  inscribed,  and  of  a  circumscribed 
equilateral  triangle. 

(Prove,  by  aid  of  Exs.  8,  10,  and  11,  p.  198,  that  the  product  of  the 
areas  of  the  inscribed  and  circumscribed  equilateral  A  is  equal  to  the 
square  of  the  area  of  the  regular  hexagon.) 

26.  If  the  diagonals  AC  and  BE  of  regular  pentagon  ABODE 
intersect  at  F,  prove  BE  =  AE  -\-  AF.     (Ex.  23.) 

27.  In  the  figure  of  Prop.  IX.,  prove  that  OM  is  the  side  of  a 
regular  pentagon  inscribed  in  a  circle  which  is  circumscribed  about 
triangle  OBM. 

(ZO^M  =  36°.) 

28.  The  area  of  the  square  inscribed  in  a  sector 
whose  central  angle  is  a  right  angle  is  equal  to  one- 
half  the  square  of  the  radius. 

(To  prove  area  ODCE  =  ^  6C^.) 

29.  The  square  inscribed  in  a  semicircle  is 
equivalent  to  two-fifths  of  the  square  inscribed 
in  the  entire  circle. 

(By  Ex.  9,  p.  198,  the  area  of  the  square  in- 
scribed in  the  entire  O  is  2  OB^ ;  we  then  have 
to  prove  area  ABCD  =  ^oi2  OB^  =  4  OB^.) 

30.  The  diagonals  AC,  BD,  CE,  DF,  FA, 
and  FB,  of  regular  hexagon  ABCDEF,  form 
a  regular  hexagon  whose  area  is  equal  to  one- 
third  the  area  of  ABCDEF. 

(The  apothem  of  GHKLMN  is  equal  to  the 
apothem  of  A  ACE,  which  may  be  found  by 
Ex.  8,  p.  198.) 


200  PLANE  GEOMETRY.  — BOOK  V. 

MEASUREMENT  OF  THE   CIRCLE. 
Prop.  XI.     Theorem. 

363.  If  a  regular  polygon  he  inscribed  in,  or  circumscribed 
about,  a  circle,  and  the  number  of  its  sides  be  indefinitely 
increased, 

I.    Its  perimeter  approaches  the  circumference  as  a  limit. 

II.   Its  area  approaches  the  area  of  the  circle  as  a  limit. 


— ^ 


Given  p  and  P  the  perimeters,  and  Tc  and  K  the  areas, 
of  two  regular  polygons  of  the  same  number  of  sides  respec- 
tively inscribed  in,  and  circumscribed  about,  a  O. 

Let  C  denote  the  circumference,  and  S  the  area,  of  the  ©. 

I.  To  Prove  that,  if  the  number  of  sides  of  the  polygons 
be  indefinitely  increased,  P  and  p  approach  the  limit  C. 

Proof.  Let  A'B'  be  a  side  of  the  polygon  whose  perimeter 
is  P,  and  draw  radius  OF  to  its  point  of  contact. 

Also,  draw  lines  OA'  and  OB'  cutting  the  circumference 
at  A  and  B,  respectively,  and  chord  AB. 

Then,  AB  is  a  side  of  the  polygon  whose  perimeter  is  p. 

(§  342) 

Now  the  two  polygons  are  similar.  (§  347) 

.-.  P:p=OA':  OF.  (§348) 

...  P-p:p=OA'-OF:OF.  (?) 

.-.  (P-p)  X  OF=p  X  (OA  -  OF).  (?) 


MEASUREMENT   OF   THE   CIRCLE.  201 

But  p  is  always  <  the  circumference  of  the  O.         (Ax.  4) 

Also,  OA'  -  OF  is  <  A'F.  (§  62) 

'-'  P-P<^y^A'F.  (1) 

Now,  if  the  number  of  sides  of  each  polygon  be  indefi- 
nitely increased,  the  polygons  continuing  to  have  the  same 
number  of  sides,  the  length  of  each  side  will  be  indefinitely 
diminished,  and  A'F  will  approach  the  limit  0. 

Then,  by  (1),  since  -—  is  a  constant,  P  —  p  will  approach 
the  limit  0.  ^^ 

But  the  circumference  of  the  O  is  <  the  perimeter  of 
the  circumscribed  polygon ;  *  and  it  is  >  the  perimeter  of 
the  inscribed  polygon.  (Ax.  4) 

Then  the  difference  between  each  perimeter  and  the  cir- 
cumference, OT  P—  C  and  C  —  p,  will  approach  the  limit  0. 

Therefore,  P  and  p  will  each  approach  the  limit  C. 

II.   To  Prove  that  K  and  k  approach  the  limit  S. 

Proof.     Since  the  given  polygons  are  similar, 

K-.k^OA^iOF".  (§349) 

.-.   K-k:k=aA^ -OF^-.OF".  (?) 

.'.  {K-k)x  OF'  =  kx  (OA''  -  OF').  (?) 

.-.  K-k  =  ^^xiOA'''-OF')  =  -^,x^F\   (?) 
OF  OF- 

Now,  if  the  number  of  sides  of  each  polygon  be  indefi- 
nitely increased,  the  polygons  continuing  to  have  the  same 
number  of  sides,  A'F  will  approach  the  limit  0. 

Then,    ^  x  A!F\   being  always    <  ^  x  A'F\   will 
02^  OF 

approach  the  limit  0. 

Whence,  K  —k  will  approach  the  limit  0. 
But  the  area  of  the  O  is  evidently  <  K,  and  >  k. 
Then,  K—  S  and  S  —  k  will  each  approach  the  limit  0. 
Therefore,  ^and  k  will  each  approach  the  limit  S. 

*  For  a  rigorous  proof  of  this  statement,  see  Appendix,  p.  380. 


202  PLANE   GEOMETRY.  — BOOK  V. 

364.  Cor.  1.  If  a  regular  polygon  be  inscribed  in  a  circle, 
and  the  number  of  its  sides  be  indefinitely  increased,  its  apo- 
them  approaches  the  radius  of  the  circle  as  a  limit 

2.  If  a  regular  polygon  be  circumscribed  about  a  circle,  and 
the  number  of  its  sides  be  indefiyiitely  increased,  its  radius 
approaches  the  radius  of  the  circle  as  a  limit. 

Prop.  XII.    Theorem. 

365.  TJie  circumferences  of  two  circles  are  to  each  other  as 
their  radii. 


Given  C  and  C  the  circumferences  of  two  (D  whose  radii 
are  M  and  M',  respectively. 

To  Prove  |  =  f 

Proof.  Inscribe  in  the  ©  regular  polygons  of  the  same 
number  of  sides;  P  and  P'  being  the  perimeters  of  the 
polygons  inscribed  in  (D  whose  radii  are  R  and  B',  re- 
spectively. 

.-.  P:P'  =  R:B'.  (§348) 

.'.  PxB'  =  P'  xE.  (?) 

Now  let  the  number  of  sides  of  each  inscribed  polygon  be 
indefinitely  increased,  the  two  polygons  continuing  to  have 
the  same  number  of  sides. 

Then,     P  x  B'  will  approach  the  limit  C  x  B', 
and  P'  X  B  will  approach  the  limit  C  x  B.  (§  363, 1.) 

By  the  Theorem  of  Limits,  these  limits  are  equal.         (?) 

.-.  CxB'=C'xB,ov  ^  =  ^.  (§  234) 


MEASUREMENT  OF  THE   CIRCLE.  203 

366.   Cor.  I.   Multiplying  the  terms  of  the  ratio  —^  by  2, 
we  have 

C      2R 


C     2  R' 
Now  let  D  and  D'  denote  the  diameters  of  the  ©  whose 
radii  are  B  and  E'j  respectively. 

That  is,  the  circumferences  of  two  circles  are  to  each  other  as 
their  diameters. 

367.  Cor.  n.   The  proportion  (1)  of  §  366  may  be  written 

§  =  §•  (§235) 

That  is,  the  ratio  of  the  circumference  of  a  circle  to  its 
diameter  has  the  same  value  for  every  circle. 

This  constant  value  is  denoted  by  the  symbol  tt. 

.-.  %=^.  (1) 

It  is  shown  by  methods  of  higher  mathematics  that  the 
ratio  TT  is  incommensurable ;  hence,  its  numerical  value  can 
only  be  obtained  approximately. 

Its  value  to  the  nearest  fourth  decimal  place  is  3.1416. 

368.  Cor.  m.     Equation  (1)  of  §  367  gives 

G  =  7rD. 

That  is,  the  circumference  of  a  circle  is  equal  to  its  diameter 
multiplied  by  tr. 

We  also  have  0  =  2  irR. 

That  is,  the  circumference  of  a  circle  is  equal  to  its  radius 
multiplied  by  2Tr.  ^ 

369.  Def.  In  circles  of  different  radii,  similar  arcs,  simi- 
lar segments,  and  similar  sector's  are  those  which  correspond 
to  equal  central  angles. 


204  PLANE   GEOMETRY.  — BOOK  V. 


Prop.  XIII.     Theorem. 

370.    The  area  of  a  circle  is  equal  to  one-half  the  product 
of  its  circumference  and  radius. 


Given  R  the  radius,  C  the  circumference,  and  S  the  area, 
of  a  O. 

To  Prove  S  =  iCxR. 

Proof.     Circumscribe  a  regular  polygon  about  the  O. 

Let  P  denote  its  perimeter,  and  K  its  area. 

Then  since  the  apothem  of  the  polygon  is  R, 

K=^PxR.  (§350) 

Now  let  the  number  of  sides  of  the  circumscribed  polygon 
be  indefinitely  increased. 

Then,  K  will  approach  the  limit  S, 

and        \PxR  will  approach  the  limit  ^0  x  R.       (§  363) 

By  the  Theorem  of  Limits,  these  limits  are  equal.         (?) 
.-.  S  =  iCxR. 

371.  Cor.  I.    We  have  0=2TrR.  (§  368) 

.-.  S  =  'kRxR  =  ttR\ 

That  is,  the  area  of  a  circle  is  equal  to  the  square  of  its 
radius  multiplied  by  tt. 

Again,  S  =  ^tt  x  4.R' =  ^tt  x  (2 R)\ 

Now  let  D  denote  the  diameter  of  the  O. 
.-.  S  =  \^L^. 

That  is,  the  area  of  a  circle  is  equal  to  the  square  of  its 
diameter  multiplied  by  Jtt. 


MEASUREMENT  OF  THE   CIRCLE.  205 

372.  Cor.  II  Let  S  and  S'  denote  the  areas  of  two  (D 
whose  radii  are  R  and  B',  and  diameters  D  and  D',  respect- 
ively. 

S  ^ttR'  ^  R^ 
' '  S'     ttR'^     R'^' 

That  is,  the  areas  of  two  circles  are  to  each  other  as  the 
squares  of  their  radii,  or  as  the  squares  of  their  diameters. 

373.  Cor.  III.  Tlie  area  of  a  sector  is  equal  to  one-half  the 
product  of  its  arc  and  radius. 

Given  s  and  c  the  area  and  arc,  respectively,  of  a  sector 
of  a  O  whose  area,  circumference,  and  radius  are  S,  G,  and 
R,  respectively. 

To  Prove  s  =  \cx  R. 

Proof.  A  sector  is  the  same  part  of  the  O  that  its  arc  is 
of  the  circumference. 

But,  1=*^-  (§^^^) 

.'.  s  =  \cx  R. 

374.  Cor.  rV.  Since  similar  sectors  are  like  parts  of  the 
(D  to  which  they  belong  (§  369),  it  follows  that 

Similar  sectors  are  to  each  other  as  the  squares  of  their 
radii. 

EXERCISES. 

31.  Find  the  circumference  and  area  of  a  circle  whose  diameter 
is  5. 

32.  Find  the  radius  and  area  of  a  circle  whose  circumference  is 

25  TT. 

33.  Find  the  diameter  and  circumference  of  a  circle  whose  area 
is  289  TT. 

34.  The  diameters  of  two  circles  are  64  and  88,  respectively. 
What  is  the  ratio  of  their  areas? 


206        PLANE  GEOMETRY.  — BOOK  V. 

Prop.  XIV.  Problem. 

375.  Given  p  and  P,  the  perimeters  of  a  regular  inscribed 
and  of  a  regular  circumscribed  polygon  of  the  same  number  of 
sides,  to  find  p'  and  P',  the  perimeters  of  a  regular  inscribed 
and  of  a  regular  circumscribed  polygon  having  double  the  num- 
ber of  sides. 

A'  M  FN  B' 


O 

Solution.  Let  AB  be  a  side  of  the  polygon  whose  perim- 
eter is  p,  and  draw  radius  OF  to  middle  point  of  arc  AB. 

Also,  draw  radii  OA  and  OB  cutting  the  tangent  to  the 
O  at  P  at  points  A'  and  B',  respectively ;  then,  A'B'  is  a 
side  of  the  polygon  whose  perimeter  is  P.  (§  342) 

Draw  chords  AF  and  BF;  also,  draw  AM  and  BN  tan- 
gents to  the  O  at  ^  and  B,  meeting  A'B'  at  M  and  JV,  re- 
spectively. 

Then  AF  and  MW  are  sides  of  the  polygons  whose 
perimeters  are  p'  and  P',  respectively.  (§  344) 

Hence,  if  n  denotes  the  number  of  sides  of  the  polygons 
whose  perimeters  are  p  and  P,  and  therefore  2  n  the  number 
of  sides  of  the  polygons  whose  perimeters  are  p'  and  P',  we 
have 

AB=^,  A'B'  =  -,  AF=^,  and  MN=^.        (1) 

n  n  2n  2n 

Draw  line  0M\  then  OJf  bisects  Z  A! OF,  (§  175) 

.-.  A!M'.  MF=  OA' :  OF.  (§  249) 

But  OA'  and  OF  are  the  radii  of  the  polygons  whose 
perimeters  are  P  and  p,  respectively. 

.-.  P:p=OA':OF.  (§348) 


MEASUREMENT  OF  THE   CIRCLE.  207 

.-.  P:p  =  A'M:MF.  (?) 

.-.  P+p:p  =  A'M-\-MF:MF.  (?) 

Or  P+p^A^F^^A^B' 

p         MF     ^MN 

An 
Clearing  of  fractions, 

P'(P  +  p)  =  2Pxp. 

P  +  P 
Again,  in  isosceles  A  ABF  and  AFM, 

Z  ABF  =  Z  AFM.  (§  §  193,  197) 

Therefore,  A  ABF  and  AFM  are  similar.  (§  255) 

AF^MF 

' '  AB     AF' 


(2) 


(?) 


.-.  AF^  =  ABxMF.  (?) 

.-.  p'^=p  X  p. 

.-.  p'=v5irp.  (3) 

Prop.  XV,     Problem. 

376.    To  compute  an  approximate  value  of  it  (§  367). 

Solution.  If  the  diameter  of  a  O  is  1,  the  side  of  an 
inscribed  square  is  iV2  (§  352);  hence,  its  perimeter  is 
2^2. 

Again,  the  side  of  a  circumscribed  square  is  equal  to  the 
diameter  of  the  0 ;  hence,  its  perimeter  is  4. 

We  then  put  in  equation  (2),  Prop.  XIV., 
P  =  4,  and  i)  =  2  V2  =  2.82843. 


208 


PLANE   GEOMETRY.  — BOOK   V. 


2Pxp 


3.31371. 


p+p 

We  then  put  in  equation  (3),  Prop.  XIV., 
p  =  2.82843,  and  P'  =  3.31371. 


.-.  y=Vi>xP'  =  3.06147. 
These  are  the  perimeters  of  the  regular  circumscribed 
and  inscribed  octagons,  respectively. 

Repeating  the  operation  with  these  values,  we  put  in  (2), 
P=  3.31371,  and  p  =  3.06147. 
2Pxp 


P' 


3.18260. 


P  +  P 
We  then  put  in  (3),  p  =  3.06147  and  P'  =  3.18260. 

.-.  p'=^pxP'  =  3.12145. 
These  are,  respectively,  the  perimeters  of  the  regular  cir- 
cumscribed and  inscribed  polygons  of  sixteen  sides. 
In  this  way,  we  form  the  following  table : 


No.   OF 

Perimetee  of 

Perimeter  of 

Sides. 

Reg.  CiRC.  Polygon. 

Eeg.  Insc.  Polygon. 

4 

4. 

2.82843 

8 

3.31371 

3.06147 

16 

3.18260 

3.12145 

32 

3.15172 

3.13655 

64 

3.14412 

3.14033 

128 

3.14222 

3.14128 

256 

3.14175 

3.14151 

512 

3.14163 

3.14167 

The  last  result  shows  that  the  circumference  of  a  O 
whose  diameter  is  1  is  >  3114157,  and  <  3.14163. 

Hence,  an  approximate  value  of  tt  is  3.1416,  correct  to  the 
fourth  decimal  place. 

Note.    The  value  of  tt  to  fourteen  decimal  places  is 
3.14159265358979. 


MEASUREMENT  OF  THE   CIRCLE.  209 

EXERCISES. 

35.  The  area  of  a  circle  is  equal  to  four  times  the  area  of  the 
circle  described  upon  its  radius  as  a  diameter, 

36.  The  area  of  one  circle  is  2|  times  the  area  of  another.  If  the 
radius  of  the  first  is  15,  what  is  the  radius  of  the  second  ? 

37.  The  radii  of  three  circles  are  3,  4,  and  12,  respectively.  What 
is  the  radius  of  a  circle  equivalent  to  their  sum  ? 

38.  Find  the  radius  of  a  circle  whose  area  is  one-half  the  area  of  a 
circle  whose  radius  is  9. 

39.  If  the  diameter  of  a  circle  is  48,  what  is  the  length  of  an  arc 
of  85°  ? 

40.  If  the  radius  of  a  circle  is  3  V3,  what  is  the  area  of  a  sector 
whose  central  angle  is  152°  ? 

41.  If  the  radius  of  a  circle  is  4,  what  is  the  area  of  a  segment 
whose  arc  is  120°  ?     (tt  =  3.1416.) 

(Subtract  from  the  area  of  the  sector  whose  central  A  is  120°,  the 
area  of  the  isosceles  A  whose  sides  are  radii  and  whose  base  is  the 
chord  of  the  segment. ) 

42.  Find  the  area  of  the  circle  inscribed  in  a  square  whose  area 
is  13. 

43.  Find  the  area  of  the  square  inscribed  in  a  circle  whose  area 
is  196  TT. 

44.  If  the  apothem  of  a  regular  hexagon  is  6,  what  is  the  area  of 
its  circumscribed  circle  ? 

45.  If  the  length  of  a  quadrant  is  1,  what  is  the  diameter  of  the 
circle?     (7r  =  3.1416.) 

46.  The  length  of  the  arc  subtended  by  a  side  of  a  regular  inscribed 
dodecagon  is  |  tt.     What  is  the  area  of  the  circle  ? 

47.  The  perimeter  of  a  regular  hexagon  circumscribed  about  a 
circle  is  12  V3.     What  is  the  circumference  of  the  circle  ? 

48.  The  area  of  a  regular  hexagon  inscribed  in  a  circle  is  24  VS. 
What  is  the  area  of  the  circle  ? 

49.  The  side  of  an  equilateral  triangle  is  6.  Find  the  areas  of  its 
inscribed  and  circumscribed  circles. 

50.  The  side  of  a  square  is  8.  Find  the  circumferences  of  its 
inscribed  and  circumscribed  circles. 

51.  Find  the  area  of  a  segment  having  for  its  chord  a  side  of  a 
regular  inscribed  hexagon,  if  the  radius  of  the  circle  is  10.    (tt =3. 1416.) 


210  PLANE   GEOMETRY.  — BOOK   V. 

52.  A  circular  grass-plot,  100  ft.  in  diameter,  is  surrounded  by  a 
walk  4  ft.  wide.     Find  the  area  of  the  walk. 

53.  Two  plots  of  ground,  one  a  square  and  the  other  a  circle,  con- 
tain each  70686  sq.  ft.  How  much  longer  is  the  perimeter  of  the  square 
than  the  circumference  of  the  circle  ?     (tt  =  3.1416.) 

54.  A  wheel  revolves  55  times  in  travelling  ft.  What  is  its 
diameter  in  inches  ? 

If  r  represents  the  radius,  a  the  apothem,  s  the  side,  and  k  the  area, 
prove  that 

55.  In  a  regular  octagon, 

s  =  r  V2-  \/2,  a  =  i  r  V2-I-  V2,  and  A;  =  2  r^  ^/2.     (§  375) 

56.  In  a  regular  dodecagon, 

s  =  r  V2-  V3,  a  =  irV2+  V3,  and  k  =  Sr^. 

57.  In  a  regular  octagon, 

s  =  2aiV2-  l),r  =  aV4:-2V2,  andk  =  8a^(V2-  1). 

58.  In  a  regular  dodecagon, 

s  =  2  a  (2  -  V3),  r  =  2  a  V2  -  Va,  and  A;  =  12  a^  (2  -  VS). 

59.  In  a  regular  decagon,  a  =  \r  VlO  4-  2  Vb.    (§  359.) 
(Find  the  apothem  by  §  273.) 

60.  What  is  the  number  of  degrees  in  an  arc  whose  length  is  equal 
to  that  of  the  radius  of  the  circle  ?     (tt  =  3.1416.) 

(Represent  the  number  of  degrees  by  x.) 

61.  Find  the  side  of  a  square  equivalent  to  a  circle  whose  diameter 
is  3.     (Tr  =  3.1416.) 

62.  Find  the  radius  of  a  circle  equivalent  to  a  square  whose  side 
is  10.     (7r  =  3.1416.) 

63.  Given  one  side  of  a  regular  hexagon,  to  construct  the  hexagon. 

64.  Given  one  side  of  a  regular  pentagon,  to  construct  the  pentagon. 
(Draw  a  O  of  any  convenient  radius,  and  construct  a  side  of  a 

regular  inscribed  pentagon. ) 

65.  In  a  given  square,  to  inscribe  a  regular  octagon. 

(Divide  the  angular  magnitude  about  the  centre  of  the  square  into 
eight  equal  parts.) 

66.  In  a  given  equilateral  triangle  to  inscribe  a  regular  hexagon. 

67.  In  a  given  sector  whose  central  angle  is  a  right  angle,  to 
inscribe  a  square. 

Note.     For  additional  exercises  on  Book  V.,  see  p.  231. 


APPENDIX   TO   PLANE   GEOMETRY. 


MAXIMA  AND  MINIMA  OF  PLANE  FIGURES. 


Prop.  I.     Theorem. 

377.    Of  all  triangles  formed  with  two  given  sides,  that  in 
which  these  sides  are  perpendicular  is  the  maximum. 


Given,  in  A  ABC  and  A'BC,  AB  =  A'B,  and  AB  ±  BG. 
To  Prove  area  ABC  >  area  A' BO. 

Proof.     Draw  A'D  ±  BO-,  then, 
A'B>A'D. 
.'.  AB>A'D. 
Multiplying  both  members  of  (1)  by  ^BO, 
iBOxAB>iBOxA'D. 
.'.  area  J.5(7  >  area  J^'^C 


(§46) 

(1) 


(§  312) 

378.  Def.     Two  figures  are  said  to  be  isoperimetric  when 
they  have  equal  perimeters. 

211 


212 


PLANE   GEOMETRY. — APPENDIX. 


Prop.  II.     Theorem. 

379.    Of  isopenmetric  triangles  having  the  same  base,  that 
which  is  isosceles  is  the  maximum. 


Given  ABC  and  A'BC  isoperimetric  A,  having  the  same 
base  BC,  and  A  ABC  isosceles. 
To  Prove  area  ABC  >  area  A'BC. 

Proof.     Produce  BA  to  D,  making  AD  =  AB,  and  draw 
line  CD. 

Then,  /.  BCD  is  a  rt.  Z  ;  for  it  can  be  inscribed  in  a  semi- 
circle, whose  centre  is  A  and  radius  AB.  (§  195) 
Draw  lines  AF  and  A'G  1.  to  (72);  take  point  E  on  CD 
so  that  A:E  =  A"C,  and  draw  line  BE. 
Then  since  A  ABC  and  A'BC  are  isoperimetric, 
AB^AC  =  A'B  +  A'C  =  A'B  +  A'E. 
.-.  A'B  +  A'E  =AB-\-AD=  BD. 
But,                  A'B  +  ^'^  >  BE. 
.'.  BD>BE. 
.'.  CD>CE. 

Now  AF  and  J.'6r  are  the  Js  from  the  vertices  to  the  bases 
of  isosceles  AACD  and  A'CE,  respectively. 
.-.  CF=^  CD,  and  CG  =  i  CE. 
.-.  CF>CG. 
Multiplying  both  members  of  (1)  by  i  BG, 
iBCx  CF>iBCx  CG. 
.'.  Sivea,  ABC >2iTea,  A'BC. 


(Ax.  4) 

(§51) 


(§94) 
(1) 


(?) 


MAXIMA  AND  MINIMA  OF  PLANE  FIGURES.       213 

380.  Cor.  Of  isoperimetric  triangles,  that  which  is  equi- 
lateral is  the  maximum. 

For  if  the  maximum  A  is  not  isosceles  when  any  side  is 
taken  as  the  base,  its  area  can  be  increased  by  making  it 
isosceles.  (§  379) 

Then,  the  maximum  A  is  equilateral. 

Prop.  III.    Theorem. 

381.  Of  isoperimetric  polygons  having  the  same  number  of 
sides,  that  which  is  equilateral  is  the  maximum. 


E  D 

Given  ABODE  the  maximum  of  polygons  having  the 
given  perimeter  and  the  given  number  of  sides. 

To  Prove  ABODE  equilateral. 

Proof.     If  possible,  let  sides  AB  and  BO  be  unequal. 

Let  AB'C  be  an  isosceles  A  with  the  base  AO,  having  its 
perimeter  equal  to  that  of  A  ABO. 

.-.  area  AB'O  >  area  ABO.  (§  379) 

Adding  area  AODE  to  both  members, 

area  AB'ODE  >  area  ABODE. 

But  this  is  impossible ;  for,  by  hyp.,  ABODE  is  the  maxi- 
mum of  polygons  having  the  given  perimeter. 
Hence,  AB  and  BO  cannot  be  unequal. 
In  like  manner  we  have 

BO  =  OD  =  DE,  etc. 
Then,  ABODE  is  equilateral. 


214 


PLANE  GEOMETRY.  —  APPENDIX. 


Prop.  IV.     Theorem. 

382.  Of  isoperimetric  equilateral  polygons  having  the  same 
number  of  sides,  that  which  is  equiangular  is  the  maximum. 

Given  AB,  BC,  and  CD  any  three  consecutive  sides  of 
the  maximum  of  isoperimetric  equilateral  polygons  having 
the  same  number  of  sides. 

To  Prove  Z  ABO  =  Z  BCD. 

Proof.     There  may  be  three  cases : 

1.  ABC  +  BCD  =  180°.     (Fig.  1.) 

2.  ABC  +  BCD  >  180°.     (Fig.  2.) 

3.  ABC  +  BCD  <1S0°.     (Fig.  3.) 

H 


Fig.  1.  Fig.  2.  Fig.  3. 

If  possible,  let  Z  ABC  be  >Z  BCD,  and  draw  line  AD. 
In  Fig.  1. 

Let  E  be  the  middle  point  of  BC]  and  draw  line  EF, 
meeting  AB  produciBd  at  F,  making  EF  =  BE. 
Produce  FE  to  meet  CD  at  G. 
Then  in  A  BEF  and  CEG,  by  hyp.,  BE  =  CE. 

Also,  Z  BEF  =  Z  CEO.  (?) 


MAXIMA  AND  MINIMA  OP  PLANE  FIGURES.       215 

And,  ZEBF=ZC, 

for  each  is  the  supplement  of  Z  B.  (§  33^  2) 

.-.  ABEF=ACEG.  (§§86,68) 

.-.  BE  =EF=CE=  EG,  and  BF=  CG.     (§  66) 
In  Fig.  2. 
Produce  AB  and  DC  to  meet  at  H. 
'  Since,  by  hyp.,  Z  ABC  >  Z  BCD,  Z  CBH<  Z  J5(7^. 

.'.BH>GH.  (§99) 

Lay  off,  on  BH,  FH  =  CH;  and  on  DH,  GH=BH;  and 
draw  line  FG  cutting  BC  Sit  E. 

.-.  A  FGH  =  A  BCH.  (§  63) 

.-.  ZCBH=ZFGH.  (§  66) 

Then,  in  A  BEF  and  CEG,  Z  EBF  =  Z  CGE. 
Also,  Z  5^i^  =  Z  (7J5;G^.  (?) 

And  BF=:CG, 

since  BF=BH-  FH,  and  CG  =  GH  -  CH. 

.'.  A  BEF=  A  CEG.  (§§  86,  68) 

.-.  BE  =  CJ^;  and  jE;i^=  EG.  (§  66) 

/w  iFYgr.  3. 
Produce  BA  and  (7Z>  to  meet  at  K. 
Since,  by  hyp.,  Z  ^^(7  >  Z  ^(7D,  (7/r  >  ^JT.  (?) 

Lay  off,  on  ir^  produced,  FK=  CK;  and  on  CK,  GK=BK; 
and  draw  line  FG  cutting  BC  at  E. 

.'.  ABCK=AFGK.  (?) 

.-.  ZF=ZC.  (?) 

Then,  in  A  ^^i^  and  O^G^,  ZF==ZG. 
Also,  Z  J5^i^  =  Z  (7^(^.  (?) 

And  BF=CG, 

since  JBi?^  =  2^7f  -  BK,  and  CG  =  CK  -  GK. 

.'.  ABEF^ACEG.  (?) 

.-.  5^;  =  CE  and  ^F  =  ^^  (?) 


216  PLANE   GEOMETRY.  — APPENDIX. 

Then  since,  in  either  figure,  BO -\- CG  =  BF  +  FG,  and 
A  BEF=  A  CEG,  quadrilateral  AFGD  is  isoperimetric  with, 
and  =0=  to,  quadrilateral  ABCD. 

Calling  the  remainder  of  the  given  polygon  P,  it  follows 
that  the  polygon  composed  of  AFGD  and  P  is  isoperimetric 
with,  and  =0=  to,  the  polygon  composed  of  ABCD  and  P; 
that  is,  the  given  polygon. 

Then  the  polygon  composed  of  AFGD  and  P  must  be 
the  maximum  of  polygons  having  the  given  perimeter  and 
the  given  number  of  sides. 

Hence,  the  polygon  composed  of  AFGD  and  P  is  equi- 
lateral. (§  381) 

But  this  is  impossible,  since  AF  is  >DG. 

Hence,  Z  ABC  cannot  be  >  Z  BCD. 

In  like  manner,  Z  ABC  cannot  be  <  Z  BCD. 

.-.  ZABC  =  ZBCD. 

Note.  The  case  of  triangles  was  considered  in  §  380.  Fig.  3  also 
provides  for  the  case  of  triangles  by  supposing  B  and  K  to  coincide 
with  A.    In  the  case  of  quadrilaterals,  P  =  0. 

383.  Cor.  Of  isoperimetric  polygons  having  the  same  num- 
ber of  sides,  that  which  is  regular  is  the  maximum. 

Prop.  V.     Theorem. 

384.  Of  two  isoperimetric  regular  polygons,  that  which  has 
the  greater  number  of  sides  has  the  greater  area. 


Given  ABC  an  equilateral  A,  and  M  an  isoperimetric 
square. 
To  Prove  area  M  >  area  ABC. 


SYMMETRICAL  FIGURES.  217 

Proof.     Let  D  be  any  point  in  side  AB  of  A  ABC, 
Draw  line  DC,   and  construct  isosceles  A  CDE  isoperi- 
metric  with  A  BCD,  CD  being  its  base. 

.-.  area  CDE  >  area  BCD.  (§  379) 

.-.  area  A  DEC  >  area  ABC 
But,  since  ADEC  and  M  are  isoperimetric, 

area  M  >  area  ADEC  (§  381) 

.-.  area  illf  >  area  ^5C. 
In  like  manner,  we  may  prove  the  area  of  a  regular  pen- 
tagon greater  than  that  of  an  isoperimetric  square ;  etc. 

385.  Cor.  The  area  of  a  circle  is  greater  than  the  area  of 
any  polygon  having  an  equal  perimeter. 

SYMMETRICAL  FIGURES. 
DEFINITIONS. 

386.  Two  points  are  said  to  be  symmetrical  with  respect 
to  a  third,  called  the  centre  of  symmetry,  when  the  latter 
bisects  the  straight  line  which  joins  them. 

Thus,  if  0  is  the  middle  point  of  straight  line  AB,  points 
A  and  B  are  symmetrical  with  respect  to  0  ^  0  b 
as  a  centre.  ' ^ ' 

387.  Two  points  are  said  to  be  symmetrical  with  respect 
to  a  straight  line,  called  the  axis  of  sym- 
metry, when  the  latter  bisects  at  right 
angles    the   straight   line   which   joins 
them. 

Thus,  if  line  CD  bisects  line  AB  at     G 

right  angles,  points  A  and  B  are  sym- 
metrical with  respect  to  CD  as  an 
axis.  h 

388.  Two  figures  are  said  to  be  symmetrical  with  respect 
to  a  centre,  or  with  respect  to  an  axis,  when  to  every  point 
of  one  there  corresponds  a  symmetrical  point  in  the  other. 


D 


218 


PLANE  GEOMETRY.— APPENDIX. 


389.  Thus,  if  to  every  point  of  triangle 
ABC  there  corresponds  a  symmetrical 
point  of  triangle  A'B'O',  with  respect  to 
centre  0,  triangle  A'B'O  is  symmetrical 
to  triangle  ABC  with  respect  to  centre  0. 

Again,  if  to  every  point  of  triangle  ABC 
there  corresponds  a  symmetrical  point  of 
triangle  A'B'C,  with  respect  to  axis  DE, 
triangle  A'B'C  is  symmetrical  to  tri- 
angle ABC  with  respect  to  axis  DE. 

390.  A  figure  is  said  to  be  symmet- 
rical with  respect  to  a  centre  when 
every  straight  line  drawn  through  the 
centre  cuts  the  figure  in  two  points 
which  are  symmetrical  with  respect  to 
that  centre. 


^■^w 


-E 


391.  A  figure  is  said  to  be  symmetrical  with  respect  to 
an  axis  when  it  divides  it  into  two  figures  which  are  sym- 
metrical with  respect  to  that  axis. 

Pkop.  VI.     Theorem. 

392.  Two  straight  lines  which  are  symmetrical  ivith  respect 
to  a  centre  are  equal  and  parallel. 


.,""0 -, 


Given  str.  lines  AB  and  AB'  symmetrical  with  respect  to 
centre  0. 

To  Prove  AB  and  AB'  equal  and  ||. 

Proof.    Draw  lines  AA,  BB',  AB\  and  AB. 


SYMMETRICAL  FIGURES. 


219 


Then,  0  bisects  AA'  and  BB\ 

Therefore,  AB'AB  is  a  O. 

Whence,  AB  and  A'B'  are  equal  and  ||. 

Prop.  VII.     Theorem. 


(§  386) 

(§  112) 

(?) 


393.  If  a  figure  is  symmetrical  with  respect  to  two  axes 
at  right  angles  to  each  other,  it  is  symmetrical  with  respect 
to  their  intersection  as  a  centre. 

Y 


p 

r 

A 

J^ 

/....., 

\p 

R    y^ 

D 

H 

> 

Q   / 

< 

0 

E 

Q 

F 

Given  figure  AE  symmetrical  with  respect  to  axes  XX' 
and  YY',  intersecting  each  other  at  rt.  A  at  0. 

To  Prove  AE  symmetrical  with  respect  to  0  as  a  centre. 

Proof.     Let   P  be   any   point  in  the  perimeter  of  AE. 

Draw  line  PQ  ±  XX',  and  line  PE  ±  YY'. 

Produce  PQ  and  PR  to  meet  the  perimeter  of  AE  at  P 
and  P",  respectively,  and  draw  lines  QR,  OP,  and  OP". 

Then  since  AE  is  symmetrical  with  respect  to  XX', 

PQ  =  P^Q.  (§  387) 

But  PQ=  OR;  whence,  OR  is  equal  and  ||  to  P'Q. 

Therefore,  OP'QR  is  a  O.  (?) 

Whence,  QR  is  equal  and  ||  to  OP'.  (?) 

In  like  manner,  we  may  prove  OP"RQ  a  O;  and  there- 
fore QR  equal  and  ||  to  OP". 

Then  since  both  OP'  and  OP"  are  equal  and  ||  to  QR, 
P'OP"  is  a  str.  line  which  is  bisected  at  0. 

That  is,  every  str.  line  drawn  through  0  is  bisected  at 
that  point,  and  hence  AE  is  symmetrical  with  respect  to  0 
as  a  centre.  (§  390) 


220 


PLANE   GEOMETRY.  —  APPENDIX. 


ADDITIONAL   EXERCISES. 
BOOK  I. 

1.  Every  point  within  an  angle,  and  not  in  the  bisector,  is  un- 
equally distant  from  the  sides  of  the  angle. 

(Prove  by  Beductio  ad  Absurdum.) 

2.  If  two  lines  are  cut  by  a  third,  and  the  sum  of  the  interior 
angles  on  the  same  side  of  the  transversal  is  less  than  two  right 
angles,  the  lines  will  meet  if  sufficiently  produced. 

(Prove  by  Beductio  ad  Absurdum.) 

3.  State  and  prove  the  converse  of  Prop.  XXXVII.,  11. 
(Prove  Z BAD  -\-ZB  =  180°.) 

4.  The  bisectors  of  the  exterior  angles  of  a  tri- 
angle form  a  triangle  whose  angles  are  respectively 
the  half-sums  of  the  angles  of  the  given  triangle 
taken  two  and  two.     (Ex.  69,  p.  67.) 

(To  prove  ZA'  =  l(Z ABC  +  Z BCA),  etc.) 

5.  If  CD  is  the  perpendicular  from  C  to  side  AB  of  triangle 
ABC,  and  CE  the  bisector  of  angle  C,  prove  ZDCJS^  equal  to  one- 
half  the  difference  of  angles  A  and  B. 

6.  If  E,  F,  G,  and  H  are  the  middle  points  of  sides  AB,  BC,  CD, 
and  DA,  respectively,  of  quadrilateral  ABCD,  prove  JE^i^'G^ // a  paral- 
lelogram whose  perimeter  is  equal  to  the  sum  of  the  diagonals  of  the 
quadrilateral.     (§  130.) 

7.  The  lines  joining  the  middle  points  of  the  opposite  sides  of  a 
quadrilateral  bisect  each  other.     (Ex.  6,  p.  220.) 

8.  The  lines  joining  the  middle  points 
of  the  opposite  sides  of  a  quadrilateral 
bisect  the  line  joining  the  middle  points 
of  the  diagonals, 

(EKGL  is  a  ZZ7,  and  its  diagonals 
bisect  each  other.) 

9.  The  line  joining  the  middle  points  of  the 
diagonals  of  a  trapezoid  is  parallel  to  the  bases 
and  equal  to  one-half  their  difference. 


ADDITIONAL  EXERCISES.  221 

10.  If  D  is  any  point  in  side  AC  oi  triangle  ABC,  and  E,  F,  G, 
and  ^the  middle  points  of  AD,  CD,  BC,  and  AB,  respectively,  prove 
EFGH  a,  parallelogram. 

11.  If  ^  and  G  are  the  middle  points  of  sides  AB  and  CD,  respec- 
tively, of  quadrilateral  ABCD,  and  K  and  L  the  middle  points  of 
diagonals  A  C  and  BD,  respectively,  prove  A  EKL  =  A  GKL. 

12.  If  D  and  E  are  the  middle  points  of 
sides  BC  and  AC,  respectively,  of  triangle 
ABC,  and  AD  be  produced  to  F  and  BE  to 
G,  making  DF  =  AD  and  EG  =  BE,  prove 
that  line  FG  passes  through  C,  and  is  bisected 
at  that  point. 

13.  If  D  is  the  middle  point  of  side  BC  ot  triangle  ABC,  prove 
AD<UAB  +  AC). 

(Produce  AD  to  E,  making  DE  =  AD.) 

14.  The  sum  of  the  medians  of  a  triangle  is  less  than  the  perimeter, 
and  greater  than  the  semi-perimeter  of  the  triangle. 

(Ex.  13,  p.  221,  and  Ex.  106,  p.  71.) 

15.  If  the  bisectors  of  the  interior  angle  at  C  and  the  exterior  angle 
at  B  of  triangle  ABC  meet  at  D,  prove  Z  BDC  =\/.A. 

16.  If  AD  and  BD  are  the  bisectors  of  the  exterior  angles  at  the 
extremities  of  the  hypotenuse  of  right  triangle  ABC,  and  DE  and  DF 
are  drawn  perpendicular,  respectively,  to  CA  and  CB  produced,  prove 
CEDE  a  square. 

(Z>  is  equally  distant  from  AC  and  BC.) 

17.  AD  and  BE  are  drawn  from  two  of  the  vertices  of  triangle 
ABC  to  the  opposite  sides,  making  Z  BAD  =  Z  ABE ;  if  ^2>  =  BE, 
prove  the  triangle  isosceles. 

18.  If  perpendiculars  AE,  BF,  CG,  and  DH,  be  drawn  from  the 
vertices  of  parallelogram  ABCD  to  any  line  in  its  plane,  not  inter- 
secting its  surface,  prove 

AE+CG  =  BF-^DH. 

(The  sum  of  the  bases  of  a  trapezoid  is  equal  to  twice  the  line 
joining  the  middle  points  of  the  non-parallel  sides.) 

19.  If  CD  is  the  bisector  of  angle  C  of  triangle 
ABC,  and  DF  be  drawn  parallel  to  AC  meeting 
BC  Bit  E  and  the  bisector  of  the  angle  exterior  to    g 
C  at  F,  prove  DE  =  EF. 


222  PLANE   GEOMETRY.— APPENDIX. 

20.  If  E  and  F  are  the  middle  points  of  sides  AB  and  AC,  re- 
spectively, of  triangle  ABC,  and  AD  the  perpendicular  from  ^  to  BC, 
prove  ZEDF=  ZEAF.    (Ex.  83,  p.  09.) 

21.  If  the  median  drawn  from  any  vertex  of  a  triangle  is  greater 
than,  equal  to,  or  less  than  one-half  the  opposite  side,  the  angle  at 
that  vertex  is  acute,  right,  or  obtuse,  respectively.     (§  98.) 

22.  The  number  of  diagonals  of  a  polygon  of  n  sides  is  ^  ^'^  ~  •"). 

23.  The  sum  of  the  medians  of  a  triangle  is  greater  than  three- 
fourths  the  perimeter  of  the  triangle. 

(Fig.  of  Prop.  LII.  Since  AO  =  ^  AD  and  -BO  =  f  BE,  we  have 
AB<  I  {AD  +  BE),  by  Ax.  4.) 

24.  If  the  lower  base  AD  of  trapezoid  ABCD      u  ^c 

is  double  the  upper  base  BC,  and  the  diagonals        /tN.E/'X 
intersect  at  E,  prove  CE  =  \  AC  and  BE  =  a  BD.       /  'v/'^N^X 

(Let  F  be  the  middle  point  of  DE,  and  G  of  L/g          ^\\ 

AE.)  ^^ ^° 

25.  If  0  is  the  point  of  intersection  of  the  p^ 
medians  AD  and  BE  of  equilateral  triangle  ABC, 
and  line  OF  be  drawn  parallel  to  side  AC,  meet-  > 
ing  side  BC  a,t  F,  prove  that  DF  is  equal  to  ^  .BC.  / 
(§133.)  l^ 

(Let  G  be  the  middle  point  of  Ovl.)  d  f  H    ^ 

26.  If  equiangular  triangles  be  constructed  on  the  sides  of  a  tri- 
angle, the  lines  drawn  from  their  outer  vertices  to  the  opposite  vertices 
of  the  triangle  are  equal.     (§  63.) 

27.  If  two  of  the  medians  of  a  triangle  are  equal,  the  triangle  is 
isosceles. 

(Fig.  of  Prop.  LII.    Let  AD  =  BE.) 

BOOK  IL 

28.  AB  and  AC  are  the  tangents  to  a  circle  from  point  A,  and  D 
is  any  point  in  the  smaller  of  the  arcs  subtended  by  chord  BC.  If  a 
tangent  to  the  circle  at  D  meets  AB  at  E  and  ^C  at  F,  prove  the 
perimeter  of  triangle  Ji-E'i?' constant.     (§  174.) 

29.  The  line  joining  the  middle  points  of  the  arcs  subtended  by 
sides  AB  and  AC  of  an  inscribed  triangle  ABC  cuts  AB  at  F  and  AC 
aXG.    Trove  AF=  AG. 

(Z  AFG  =  Z  AGF.) 


ADDITIONAL  EXERCISES. 


223 


30.  If  ABCD  is  a  circumscribed  quadrilateral,  prove  the  angle 
between  the  lines  joining  the  opposite  points  of  contact  equal  to 
K^+C).     (§202.) 

31.  If  sides  AB  and  EC  ot  inscribed  hexagon  ABCDEF  are 
parallel  to  sides  DE  and  EF,  respectively,  prove  side  AF parallel  to 
side  CD.     (§  172.) 

(Draw  line  CF,  and  prove  Z.  AFC  =  /.  FCD.) 

32.  If  AB  is  the  common  chord  of  two  inter- 
secting circles,  and  AC  and  AD  diameters  drawn 
from  A^  prove  that  line  CD  passes  through  B. 
(§  195.)  c 

33.  If  AB  is  a  common  exterior  tangent  to  two  circles  which  touch 
each  other  externally  at  C,  prove  Z.ACB  a  right  angle. 

(Draw  the  common  tangent  at  C,  meeting  AB  at  D.) 

34.  If  ^B  and  AC  are  the  tangents  to  a  circle  from  point  A,  and 
D  is  any  point  on  the  greater  of  the  arcs  subtended  by  chord  BC, 
prove  the  sum  of  angles  ABD  and  ACD  constant. 


35.  If  ^,  C,  B,  and  D  are  four  points  in 
a  straight  line,  B  being  between  C  and  Z>,  and 
EF  is  a  common  tangent  to  the  circles  described 
upon  AB  and  CD  as  diametei-s,  prove 

ZBAE  =  ADCF. 
(We  have  0^11  O'F.) 

36.  ABCD  is  an  inscribed  quadrilateral, 
AD  being  a  diameter  oi  the  circle.  If  0  is  the 
centre,  and  sides  AD  and  BC  produced  meet 
at  E  making  CE  —  OA,  prove 

ZAOB  =  SZCED. 
(ZAOB  is  an  ext.  Z  of  A  QBE,  and  Z BCO 
of  A  OCE.) 

37.  ABCD  is  a  quadrilateral  inscribed  in 
a  circle.  If  sides  AB  and  DC  produced  inter- 
sect at  E,  and  sides  AD  and  BC  produced 
at  F,  prove  the  bisectors  of  angles  E  and  F 
perpendicular.     (§  199.) 

(Prove  arc  HxM  +  arc  KL  =  180°.) 


224        PLANE  GEOMETRY. —APPENDIX. 

38.  If  ABCD  is  an  inscribed  quadrilateral,  and  sides  AD  and  BC 
produced  meet  at  P,  the  tangent  at  P  to  the  circle  circumscribed  about 
triangle  ABP  is  parallel  to  CD.     (§  196.) 

(Prove  Z  between  the  tangent  and  BP  equal  to  Z  PCD.) 

39.  ABCD  is  a  quadrilateral  inscribed  in  a  circle.  Another  circle 
is  described  upon  AD  as  a  chord,  meeting  AB  and  CD  at  JS  and  F, 
respectively.     Prove  chords  BC  and  ^P  parallel. 

(Vrove  Z ABC  =ZAEF.) 

40.  If  ABCDEFGH  is  an  inscribed  octagon,  the  sum  of  angles 
A^  C,  F,  and  G  is  equal  to  six  right  angles.     (§  193.) 

41.  If  the  number  of  sides  of  an  inscribed  polygon  is  even,  the 
sum  of  the  alternate  angles  is  equal  to  as  many  right  angles  as  the 
polygon  has  sides  less  two. 

(Use  same  method  of  proof  as  in  Ex.  40.) 

42.  If  a  right  triangle  has  for  its  hypotenuse  the  side  of  a  square, 
and  lies  without  the  square,  the  straight  line  drawn  from  the  centre 
of  the  square  to  the  vertex  of  the  right  angle  bisects  the  right  angle. 
(§  200.) 

43.  The  perpendiculars  from  the  vertices  of  a  triangle  to  the  oppo- 
site sides  are  the  bisectors  of  the  angles  of 

the  triangle  formed  by  joining  the  feet  of 
the  perpendiculars. 

(To  prove  AD,  BE,  and  CF  the  bisect- 
ors of  the  A  ol  A  DEF.  By  §  200,  a  O 
can  be  circumscribed  about  quadrilateral 
BDOF;  then  ZODF=ZOBF;  in  this 
way,  Z  ODF  =dO°-Z BAC.) 

Constructions. 

44.  Given  a  side,  an  adjacent  angle,  and  the  radius  of  the  circum- 
scribed circle  of  a  triangle,  to  construct  the  triangle. 

What  restriction  is  there  on  the  values  of  the  given  lines  ? 

P 

45.  To  describe  a  circle  of  given  radius  tan-       ^_^^   /'''^^N. 

gent  to  a  given  circle,  and  passing  through  a  given      Z^^,     y         L-''    A 
point  without  the  circle.  \^_^A        '^     / 

46.  To  draw  between  two  given  intersecting  lines  a  straight  line 
which  shall  be  equal  to  one  given  straight  line,  and  parallel  to  another. 

(Draw  a  ||  to  one  of  the  intersecting  lines.) 


ADDITIONAL  EXERCISES. 


225 


47.  Given  an  angle  of  a  triangle,  the  length  of  its  bisector,  and 
the  length  of  the  pei-pendicular  from  its  vertex  to  the  opposite  side, 
to  construct  the  triangle . 

(The  side  opposite  the  given  Z.  is  tangent  to  a  0  drawn  with  the 
vertex  as  a  centre,  and  with  the  ±  from  the  vertex  to  the  opposite  side 
as  a  radius.) 


48.  Given  an  angle  of  a  triangle,  and  the  segments  of  the  oppo- 
site side  made  by  the  perpendicular  from  its  vertex,  to  construct  the 
triangle.     (§  226.) 

B 


49.   To  inscribe  a  square  in  a  given  rhombus. 

(Bisect  the  A  between  diagonals  AG  and  BB.  To 
prove  EFGH  a  square,  prove  k.  OBE,  OBF,  ODG, 
and  ODH  equal  j  whence,  OE=OF=  0G=  OH.) 


50.  To  draw  a  parallel  to  side  BC  of  triangle 
ABC  meeting  AB  and  AC  in  D  and  E,  respec- 
tively, so  that  DE  may  equal  EC. 

51.  To  draw  a  parallel  to  side  BC  of  tri- 
angle ABC,  meeting  AB  and  AC  in  D  and  E, 
respectively,  so  that  DE  may  equal  the  sum  of 
BD  and  CE. 


52.  Given  an  angle  of  a  triangle,  the  length  of  the  perpendicular 
from  the  vertex  of  another  angle  to  the  opposite  side,  and  the  radius 
of  the  circumscribed  circle,  to  construct  the  triangle. 

(The  centre  of  the  circumscribed  O  is  equally  distant  from  the 
given  vertices.) 

53.  Through  a  given  point  without  a  given  circle  to  draw  a  secant 
whose  internal  and  external  segments  shall  be  equal.    (Ex.  65,  p.  103.) 


54.  Given  the  base  of  a  triangle,  an  adjacent 
angle,  and  the  sum  of  the  other  two  sides,  to  con- 
struct the  triangle. 

(Lay  off  AD  equal  to  the  sum  of  the  other  two 

•) 


E 


226  PLANE   GEOMETRY.  — APPENDIX. 

55.  Given  the  base  of  a  triangle,  an  adjacent 
acute  angle,  and  the  difference  of  the  other  two 
sides,  to  construct  the  triangle. 

What  restriction  is  there  on  the  values  of  the 
given  lines?  A- 

56.  Given  the  feet  of  the  perpendiculars  from  the  vertices  of  a 
triangle  to  the  opposite  sides,  to  construct  the  triangle.     (Ex.  43.) 

BOOK  III. 

57.  In  any  triangle,  the  product  of  any  two 

sides  is  equal  to  the  product  of  the  segments  ^^x^        x'P^"'-- 

of  the  third  side  formed  by  the  bisector  of  the  ^>K^     \     '> 

exterior  angle  at  the  opposite  vertex,  minus  ^/''^'•A^^X  / 

the  square  of  the  bisector.  ^-^ Jr — ^c 

(To    prove  AB  x  AC  =  DB  x  DC  -  AIT. 
The  work  is  carried  out  as  in  §  288;  first  prove  A  ABD  and  ACE 
similar.) 

58.  If  the  sides  of  a  triangle  are  AB  =  4,  AC  =5,  and  BC  =  0, 
find  the  length  of  the  bisector  of  the  exterior  angle  at  vertex  A. 
(§  251.) 

59.  ABC  is  an  isosceles  triangle.  If  the  perpendicular  to  AB  at 
A  meets  base  BC,  produced  if  necessary,  at  E,  and  D  is  the  middle 
point  of  BE,  prove  AB  a  mean  proportional  between  BC  and  BD. 
(Ex.  83,  p.  69.) 

(A  ABC  and  ABD  are  similar.) 

60.  If  D  and  E,  F  and  G,  and  H  and  K  are  points 
on  sides  AB,  BC,  and  CA,  respectively,  of  triangle 
ABC,  so  taken  that  AD=DE^EB,  BF=FG=GC,  N^ 
and  CH=  HK=  KA,  prove  that  lines  EF,  GH,  and 
KD,  when  produced,  form  a  triangle  equal  to  ABC. 

(By  §  248,  sides  of  I\LMN  are  ||,  respectively,  to 
sides  of  A  ^BC.) 

61.  The  square  of  the  common  tangent  to  two  circles  which  are 
tangent  to  each  other  externally  is  equal  to  4  times  the  product  of 
their  radii.     (§  273.) 

62.  The  sides  AB  and  BC  of  triangle  ABC  are  3  and  7,  respec- 
tively, and  the  length  of  the  bisector  of  the  exterior  angle  B  is  SVf. 
Find  side  AC.     (Ex.  67,  and  §  261.) 

63.  One  segment  of  a  chord  drawn  through  a  point  7  units  from 
the  centre  of  a  circle  is  4  units.  If  the  diameter  of  the  circle  is  15 
units,  what  is  the  other  segment  ?     (§  280.) 


ADDITIONAL  EXEKCISES. 


227 


64.  If  E  is  the  middle  point  of  one  of  the  parallel  sides  BC  of 
trapezoid  ABCD^  and  AE  and  DE  produced  meet  DC  and  AB  pro- 
duced at  F  and  6r,  respectively,  prove  FCr  parallel  to  AD. 

(A  ADCr  and  BEG  are  similar,  as  also  are  M^  ADF  and  CEF.) 

65.  The  perpendicular  from  the  intersection 
of  the  medians  of  a  triangle  to  any  straight  line 
in  the  plane  of  the  triangle,  not  intersecting  its 
surface,  is  equal  to  one-third  the  sum  of  the 
perpendiculars  from  the  vertices  of  the  triangle 
to  the  same  line. 

(The  sum  of  the  bases  of  a  trapezoid  is  equal 
to  twice  the  line  joining  the  middle  points  of 
the  non-parallel  sides.) 

66.  If  two  parallels  are  cut  by  three  or  more 
straight  lines  passing  through  a  common  point, 
the  corresponding  segments  are  proportional. 

CTo  prove  AE.  -  B^  -  ^^ 
(10  prove  ^,^,  -  ^,^,  -  ^^^ 

OBC,  and  OCD  are  similar,  respectively 
^  OA'B',  OB'C,  and  OCA'.) 


AOAB, 


to 


67.  State  and  prove  the  converse  of  Ex.  66. 

(Fig.  of  Ex.  66.  To  prove  that  AA',  BB\  CC,  and  DD'  pass 
through  a  common  point.  Let  AA'  and  BB'  meet  at  O,  and  draw 
OC  and  OC  ;  then  prove  4  OBC  and  OB'C  similar.) 

68.  The  non-parallel  sides  of  a  trapezoid  and  the  line  joining  the 
middle  points  of  the  parallel  sides,  if  produced,  meet  in  a  common 
point.     (Ex.  67.) 

69.  BD  is  the  perpendicular  from  the  vertex  of 
the  right  angle  to  the  hypotenuse  of  right  triangle 
ABC.  If  E  is  any  point  in  AB,  and  EF  be  drawn 
perpendicular  to  AC,  and  FG  perpendicular  to  AB, 
prove  lines  CE  and  DG  parallel. 

(A  ABC  and  AEF  are  similar.  By  §  271^  2, 
AD:CD  =  A&:  BC^,  and  AG  :  EG  =  AF"  :  EF^- ; 
AD  :  CD  =  AG -.EG.) 

70.  In  right  triangle  ABC,  BC'  =  3  AC"".  If  CD  be  drawn  from 
the  vertex  of  the  right  angle  to  the  middle  point  of  AB,  prove  ZACD 
equal  to  60°.     (Ex.  83,  p.  69.) 

(Frove  AC  =  ^  AB.) 


we  may  prove 
then,  we  have 


228 


PLANE   GEOMETRY.  —  APPENDIX. 


71.  If  i)  is  the  middle  point  of  side  BC  oi  right  triangle  ABCf  and 
BE  be  drawn  perpendicular  to  hypotenuse  AB,  prove 

AE^  -  BE^  =  AC^. 
(AE  =  AB  —  BE;  square  this  by  the  rule  of  Algebra.) 

72.  If  BE  and  CF  are  medians  drawn  from  the  extremities  of  the 
hypotenuse  of  right  triangle  ABC^  prove 

(§  272.) 


4.BE^  +  4:UF^  =  bBC^. 


73.  If  ABC  and  ADC  are  angles  inscribed 
in  a  semicircle,  and  AE  and  CF  be  drawn  per- 
pendicular to  BD  produced,  prove 

^^2  ^  ^2  ^  ^^2  ^  ^2       ^g  273.) 

74.  If  perpendiculars  PF,  PD,  and  P^be 
drawn  from  any  point  P  to  sides  AB^  BC,  and 
C^,  respectively,  of  a  triangle,  prove 

AF^  +  B&  +  OE^  =  AE^  +  :BP^  +  C&. 

(§  273.) 

75.  If  PC  is  the  hypotenuse  of  right  triangle  ABC,  prove 

{AB  +  PC  +  C^)2  =  2{AB  4-  PC)  (PC  +  CA). 
(Square  AB->rBC-\-CA  by  the  rule  of  Algebra.) 


76.    If  lines  be  drawn  from  any  point  P  to 
the  vertices  of  rectangle  ABCD,  prove 


PA^  ■\- PC  =  PB' ^  PD\ 


77.  If  AB  and  AC  are  the  equal  sides  of  an  isosceles  triangle, 
and  BD  be  drawn  perpendicular  to  AC,  prove  2  AC  x  CD  =  BC^. 

{AD  =  AC  —  CD ;  square  this  by  the  rule  of  Algebra.) 

78.  If  AD  and  BE  are  the  perpendiculars  from  vertices  A  and  P, 
respectively,  of  acute-angled  triangle  ABC  to  the  opposite  sides,  prove 

AC  X  AE  -\-  BC  X  BD  =  AB\ 

(By  §  277,  2ACxAE=  AB^  +  AC^  -  BC^  ;  and  in  like  manner 
a  value  may  be  found  for  2  PC  x  PP.) 

79.  The  sum  of  the  squares  of  the  sides  of  a  parallelogram  is  equal 
to  the  sum  of  the  squares  of  its  diagonals.  (§  279,  I.) 

80.  To  construct  a  triangle  similar  to  a  given  triangle,  having 
given  its  perimeter. 

(Divide  the  perimeter  into  parts  proportional  to  the  sides  of  the  A.) 


ADDITIONAL  EXERCISES. 


229 


81.  To  construct  a  right  triangle,  having  given  its  perimeter  and 
an  acute  angle. 

(From  any  point  in  one  side  of  the  given  Z  draw  a  ±  to  the  other 
side. ) 

82.  To  describe  a  circle  through  two 
given  points,  tangent  to  a  given  straight 
line.     (§  282.) 

(To  prove  O  draw  with  0  as  a  centre  and 
OP  as  a  radius  tangent  to  AB,  draw  BF  tan- 
gent to  the  O,  and  prove  A  OBE  =  A  OBF.) 

83.  If  A  and  B  are  points  on  either  side 
of  line  CD,  and  line  AB  cuts  CD  at  F,  find 
a  point  E  in  CD  such  that 

AE:BE  =  AF:  BF.     (§  249.) 
{EF  bisects  Z  AEB  of  A  ABE.) 


BOOK  IV. 

84.   In  the  figure  on  p.  174, 

(a)   Prove  lines  CF  and  BH  perpendicular. 

(If  CF  and  BH  meet  at  S,  Z  CSH  is  an  ext.  Z  of  A  BCS.) 

(6)   Prove  lines  AG  and  i?^  parallel. 

(c)  Prove  the  sum  of  the  perpendiculars  from  H  and  L  to  AB  pro- 
duced equal  to  AB. 

(If  J.  from  H  meets  BA  produced  at  Q,  A  AHQ  =  A  ACD.) 

(d)  Prove  triangles  AFH,  BEL,  and  CGK e?ich  equivalent  to  ABC. 
(If  AF  be  taken  as  the  base  of  A  AFH,  its  altitude  is  equal  to  CD.) 

(e)  Prove  (7,  H,  and  i  in  the  same  straight  line. 
(Prove  CH  and  CL  in  the  same'str.  line.) 

(/)  Prove  the  square  described  upon  the  sum  of  ^C  and  BG 
equivalent  to  the  square  described  upon  AB,  plus  4  times  A  ABC. 

(Square  AC  +  BC  by  the  rule  of  Algebra.) 

(g)  Prove  the  sum  of  angles  AFH,  AHF,  BEL,  and  BLE  equal 
to  a  right  angle. 

(Z  AFH^-  ZAHF=  180=  -  Z  FAH.) 

(h)  If  FN  and  EP  are  the  perpendiculars  from  F  and  E,  respec- 
tively, to  HA  and  LB  produced,  prove  triangles  AFN  and  BEP  each 
equal  to  ABC. 

(0   Prove  :^^  +  FH^  +  ^^^  =  6 .4^. 

(^L  is  the  hypotenuse  of  rt.  AELP,  and  FH  of  AFHN;  sides 
PL  and  ifiVmay  be  found  by  (^).) 


230 


PLANE   GEOMETRY.  —  APPENDIX. 


(j)   Prove  CF^  -  C£f  =  AC^  -  BC^. 

(k)  Prove  that  lines  /IL,  BH^  and  CM  meet  at  a  common  point. 
(Ex.  84,  («).) 

(Produce  DC  to  T,  making  CT=BM,  and  prove  AL,  BH,  and 
CM  the  Js  from  the  vertices  to  the  opposite  sides  of  A  ABT.) 

(I)  Prove  that  lines  HG,  LK,  and  MC  when  produced  meet  at  a 
common,  point. 

(Draw  GTand  KT,  and  prove  A  CGT  and  CKTrt.  A) 

85.    If  BE  and  CF  are  medians  drawn  from  .a 

vertices  B  and  C  of  triangle  J.J5C,  intersecting 


at  D,  prove  triangl 
lateral  AEDF. 


BCD  equivalent  to  quadri- 
area  BCF-  area  BDF.) 


(area  BCD 

86.  If  Z>  is  the  middle  point  of  side  BG  of  triangle  ABC,  E  the 
middle  point  of  AD^  F  of  BE^  and  6r  of  CF,  prove  A  ABC  equivalent 
toSA^i^G^. 

(Draw  CE ;  then,  area  ABC  =  2  area  BCE.) 

87.  If  jE"  and  J?' are  the  middle  points  of  sides  AB  and  CD,  respec- 
tively, of  parallelogram  ABCD,  and  AF  and  CE  be  drawn  intersecting 
BD  in  If  and  L,  respectively,  and  BF  and  DE  intersecting  J.C  in 
K  and  G,  respectively,  prove  GHKL  a  parallelogram  equivalent  to 
\ABCD.     (§140.) 

(If  vie  and  BD  intersect  at  3/,  AMawd.  DE  are  medians  of  A  ABD.) 

88.  Any  quadrilateral  ABCD  is  equivalent  to  a 
triangle,  two  of  whose  sides  are  equal  to  diagonals 
AC  and  BD,  respectively,  and  include  an  angle 
equal  to  either  of  the  angles  between  A  C  and  BD. 

(Produce  AC  to  F,  making  CF=AE;  and  BD 
to  G,  making  DG  =  BE.  To  prove  quadrilateral 
ABCD=o=AEFG.    ADFG<^AABC.) 

89.  If  through  any  point  E  in  diagonal  AC 
of  parallelogram  ABCD  parallels  to  AD  and 
AB  be  drawn,  meeting  AB  and  CD  in  i^  and 
H,  respectively,  and  BC  and  AD  in  G^  and  K, 
respectively,  prove  triangles  EFG  and  EIIK 
equivalent. 

90.  If  E  is  the  intersection  of  diagonals  AC  and  BD  of  a  quadri- 
lateral, and  triangles  ABE  and  Ci>^  are  equivalent,  prove  sides  AD 
and  J5C  parallel. 

(A  ^.BZ)  and  ^Ci>  are  equivalent.) 


ADDITIONAL  EXERCISES. 


231 


91.  Find  the  area  of  a  trapezoid  whose  parallel  sides  are  28  and 
36,  and  non-parallel  sides  15  and  17,  respectively. 

(By  drawing  through  one  vertex  of  the  upper  base  a  II  to  one  of  the 
non-parallel  sides,  one  Z  of  the  figure  may  be  proved  a  rt.  Z,  by  Ex. 
63,  p.  154.) 

92.  If  similar  polygons  be  described  upon  the  legs  of  a  right  triangle 
as  homologous  sides,  the  polygon  described  upon  the  hypotenuse  is 
equivalent  to  the  sum  of  the  polygons  described  upon  the  legs. 

(Find,  by  §  322,  the  ratio  of  the  area  of  the  polygon  described  upon 
each  leg  to  the  area  of  the  polygon  described  upon  the  hypotenuse.) 

93.  If  JE",  F,  G^  and  H  are  the  middle  points 
of  sides  AB,  BC,  CD,  and  DA,  respectively,  of  a 
square,  prove  that  lines  AG,  BH,  CE,  and  DF 
form  a  square  equivalent  to  \ABCD. 

(First  prove  A  ADG  =  A  ABH;  then,  by  §  85, 1, 
Z  NKL  may  be  proved  a  rt.  Z.  By  §  131,  each  side 
of  KLMN  may  be  proved  equal  to  AK.  From 
similar  A  J.HZTand  ADG,  AKmd^y  be  proved  equal  to 


AD 


94.  HE  is  any  pomt  in  side  BC  oi  parallelogram  ABCD,  and  DE 
be  drawn  meeting  AB  produced  at  F,  prove  triangles  ABE  and  CEF 
equivalent. 

(A  ABE  +  A  CDE <^  A  CDF.) 


95.  If  i>  is  a  point  in  side  AB  of  triangle  ABC, 
find  a  point  E  m  AC  such  that  triangle  ADE  shall 
be  equivalent  to  one-half  triangle  ABC.  ^ 

(ADEF^^ACEF)  D 

What  restriction  is  there  on  the  position  of  D  ?    B 


BOOK  V. 


96.  The  area  of  the  ring  included  between  two 
concentric  circles  is  equal  to  the  area  of  a  circle, 
whose  diameter  is  that  chord  of  the  outer  circle 
which  is  tangent  to  the  inner. 

(To  prove  area  of  ring  =  i  ttAC  .) 

97.  An  equilateral  polygon  circumscribed  about  a  circle  is  regular 
if  the  number  of  its  sides  is  odd.     (§  345.) 

(The  polygon  can  be  inscribed  in  a  O.) 


232 


PLANE  GEOMETRY.  —  APPENDIX. 


98.  An  equiangular  polygon  inscribed  in  a  circle  is  regular  if  the 
number  of  its  sides  is  odd.     (§  345.) 

(The  polygon  can  be  proved  equilateral.) 

99.  If  a  circle  be  circumscribed  about  a  right 
triangle,  and  on  each  of  its  legs  as  a  diameter 
a  semicircle  be  described  exterior  to  the  triangle, 
the  sum  of  the  areas  of  the  crescents  thus  formed 
is  equal  to  the  area  of  the  triangle.     (§  272.) 

(To  prove  area  AECG  -\-  area  BFCH  equal 
to  area  ABC.) 

100.  If  the  radius  of  the  circle  is  1,  the  side,  apothem,  and  diagonal 
of  a  regular  inscribed  pentagon  are,  respectively, 

I  V(10  -  2  \/5),  ^  (1  +  V5),  and  |  \/(10  +  2  V5). 
(In  Fig.  of  Prop.  IX,,  the  apothem  of  a  regular  inscribed  pentagon 
is  the  distance  from  0  to  the  foot  of  a  ±  from  B  to  OA,  and  its  side 
is  twice  this  ±.  The  diagonal  is  a  leg  of  a  rt.  A  whose  hypotenuse 
is  a  diameter,  and  whose  other  leg  is  a  side  of  a  regular  inscribed 
decagon.) 

101.  The  square  of  the  side  of  a  regular  inscribed  pentagon,  minus 
the  square  of  the  side  of  a  regular  inscribed  decagon,  is  equal  to  the 
square  of  the  radius.     (Ex.  100,  and  §  359.) 

102.  The  sum  of  the  perpendiculars  drawn  to  the  sides  of  a  regu- 
lar polygon  from  any  point  within  the  figure  is  equal  to  the  apothem 
multiplied  by  the  number  of  sides  of  the  polygon. 

(The  Js  are  the  altitudes  of  ^  which  m.ake  up  the  polygon.) 

103.  In  a  given  equilateral  triangle  to  in- 
scribe three  equal  circles,  tangent  to  each  other, 
and  each  tangent  to  one,  and  only  one,  side  of 
the  triangle. 

(By  §  174,  the  (D  touch  the  Js  at  the  same 
points.) 


104.  In  a  given  circle  to  inscribe  three 
equal  circles,  tangent  to  each  other  and  to  the 
given  circle. 


SOLID    GEOMETRY. 


Book  YI. 


LINES  AND  PLANES  IN  SPACE.     DIEDRAL 
ANGLES.     POLYEDRAL  ANGLES. 


394.  Def.  A  plane  is  said  to  be  determined  by  certain 
lines  or  points  when  one  plane,  and  only  one,  can  be  dra\fn 
through  these  lines  or  points. 


395. 

I. 

IL 

IIL 

IV. 


Prop.  I.     Theorem. 

A  plane  is  determined 

By  a  straight  line  and  a  point  without  the  line. 

By  three  points  not  in  the  same  straight  line. 

By  two  intersecting  straight  lines. 

By  two  parallel  straight  lines. 


I.   Given  point  C  without  str.  line  AB. 

To  Prove  that  a  plane  is  determined  by  AB  and  C. 

Proof.  If  any  plane  MN  be  drawn  through  AB,  it  may 
be  revolved  about  AB  as  an  axis  until  it  contains  point  C. 

Hence,  a  plane  can  be  drawn  through  line  AB  and  point 
C;  and  it  is  evident  that  but  one  such  plane  can  be  drawn. 

233 


234 


SOLID   GEOMETRY.— BOOK   VI. 


II.  Given  A,  B,  and  C  three  points  not  in  the  same  str. 
line. 

To  Prove  that  a  plane  is  determined  by  A,  B,  and  C. 

Proof.  Draw  line  AB\  then  a  plane,  and  only  one,  can 
be  drawn  through  line  AB  and  point  C. 

[A  plane  is  determined  by  a  str.  line  and  a  point  without  the  line.] 

(§  395,  I) 

Then,  a  plane,  and  only  one,  can  be  drawn  through  A,  B, 
and  C. 


III.   Given  AB  and  BC  intersecting  str.  lines. 

To  Prove  that  a  plane  is  determined  by  AB  and  BC. 

Proof.  A  plane,  and  only  one,  can  be  drawn  through 
line  AB  and  point  C. 

[A  plane  is  determined  by  a  str.  line  and  a  point  without  the  line.  ] 

(§  395,  I) 

And  since  this  plane  contains  points  B  and  C,  it  must 
contain  line  BC. 

[A  plane  is  a  surface  such  that  the  str.  line  joining  any  two  of  its 
points  lies  entirely  in  the  surface.]  (§  9) 

Then,  a  plane,  and  only  one,  can  be  drawn  through  AB 
and  BC. 


IV.   Given  lis  AB  and  CD. 

To  Prove  that  a  plane  is  determined  by  AB  and  CD. 
Proof.     The  lis  AB  and  CD  lie  in  the  same  plane. 
[Two  str.  lines  are  said  to  be  ||  when  they  lie  in  the  same  plane, 
and  cannot  meet  however  far  they  may  be  produced.]  (§  52) 


LINES  AND  PLANES  IN  SPACE. 


235 


And  only  one  plane  can  be  drawn  through  AB  and  point  C. 
[A  plane  is  determined  by  a  str.  line  and  a  point  witliout  the  line.] 

(§  395,  I) 
Then,  a  plane,  and  only  one,  can  be  drawn  through  AB 
and  CD. 

Prop.  II.     Theorem. 
396.    TJie  intersection  of  two  planes  is  a  straight  line. 


Given  line  AB  the  intersection  of  planes  MN  and  PQ. 
To  Prove  AB  a  str.  line. 

Proof.     Draw  a  str.  line  between  points  A  and  B. 
This  str.  line  lies  in  plane  MJSf,  and  also  in  plane  FQ. 
[A  plane  is  a  surface  such  that  the  str.  line  joining  any  two  of  its 
points  lies  entirely  in  the  surface.]  (§  9) 

Then  it  must  be  the  intersection  of  planes  MN  and  FQ. 
Hence,  the  line  of  intersection  AB  is  a  str.  line. 

397.  Defs.  If  a  straight  line  meets  a  plane,  the  point  of 
intersection  is  called  the  foot  of  the  line. 

A  straight  line  is  said  to  be  perpendicular  to  a  plane  when 
it  is  perpendicular  to  every  straight  line  drawn  in  the  plane 
through  its  foot. 

A  straight  line  is  said  to  be  parallel  to  a  plane  when  it 
cannot  meet  the  plane  however  far  they  may  be  produced. 

A  straight  line  which  is  neither  perpendicular  nor  parallel 
to  a  plane,  is  said  to  be  oblique  to  it. 

Two  planes  are  said  to  be  parallel  to  each  other  when  they 
cannot  meet  however  far  they  may  be  produced. 


236 


SOLID   GEOMETRY.— BOOK  VI. 


398.  Sch.     The   following  is   given   for  convenience  of 
reference : 

A  perpendicular  to  a  plane  is  perpendicular  to  every  straight 
line  drawn  in  the  plane  through  its  foot. 

Prop.  III.     Theorem. 

399.  At  a  gircen  point  in  a  plane,  one  perpendicular  to  the 
plane  cayi  he  drawn,  and  hut  one. 


Given  point  P  in  plane  MN. 

To  Prove  that  a  ±  can  be  drawn  to  MN  at  P,  and  but  one. 

Proof.  At  any  point  A  of  indefinite  str.  line  AB,  draw 
lines  AC  and  AD  ±  to  AB. 

Let  RS  be  the  plane  determined  hj  AC  and  AD. 

Let  AE  be  any  other  str.  line  drawn  through  point  A  in 
plane  RS\  and  draw  line  CD  intersecting  AC,  AE,  and  AD 
at  C,  E,  and  D,  respectively. 

Produce  BA  to  B',  making  AB'  =  AB,  and  draw  lines  BC, 
BE,  BD,  B'C,  B'E,  and  B'D. 

In  A  BCD  and  B'CD, 

CD  =  CD. 

And  since  AC  and  AD  are  ±  to  BB'  at  its  middle  point, 

BC  =  B'CsindBD  =  B'D. 

[If  a  J-  be  erected  at  the  middle  point  of  a  str.  line,  any  point  in 
the  ±  is  equally  distant  from  the  extremities  of  the  line.]         (§  41,  I) 


LINES  AND  PLANES  IN  SPACE.  237 

.-.  ABCD  =  AB'CD. 

[Two  k.  are  equal  when  the  three  sides  of  one  are  equal  respectively 
to  the  three  sides  of  the  other.]  (§  69) 

Now  revolve  A  BCD  about  CD  as  an  axis  until  it  coin- 
cides with  A  5'(7Z>. 

Then,  point  B  will  fall  at  point  B',  and  line  BE  will  coin- 
cide with  line  B'E ;  that  is,  BE  =  B'E. 

Hence,  since  points  A  and  E  are  each  equally  distant  from 
B  and  B',  line  AE  is  ±  BB'. 

[Two  points,  each  equally  distant  from  the  extremities  of  a  str.  line, 
determine  a  _L  at  its  middle  point.  ]  (§  43) 

But  AE  is  any  str.  line  drawn  through  A  in  plane  US. 

Then,  AB  is  ±  to  every  str.  line  drawn  through  A  in 
plane  BS. 

Whence,  ABh  ±  to  plane  RS. 

[A  str.  line  is  said  to  be  ±  to  a  plane  when  it  is  ±  to  every  str.  line 
drawn  in  the  plane  through  its  foot.]  (§  397) 

Now  apply  plane  BS  to  plane  MN  so  that  point  A  shall 
fall  at  point  P;  and  let  AB  take  the  position  PQ. 

Then,  PQ  will  he  ±MN. 

Hence,  a  J_  can  be  drawn  to  MN  at  P, 

If  possible,  let  PT  be  another  _L  to  plane  MN  at  P;  and 
let  the  plane  determined  by  PQ  and  PT  intersect  JOT  in 
line  HK. 

Then,  both  PQ  and  PT  are  ±  HK. 

[A  J_  to  a  plane  is  _L  to  every  str.  line  drawn  in  the  plane  through 
its  foot.]  (§398) 

But  this  is  impossible ;  for,  in  plane  HKT,  only  one  _L  can 
be  drawn  to  HK  at  P. 

[At  a  given  point  in  a  str.  line,  but  one  _L  to  the  line  can  be  drawn.] 

(§25) 
Then  only  one  ±  can  be  drawn  to  MN  at  P. 

400.  Cor.  I.  A  straight  line  perpeyidicular  to  each  of  two 
straight  lines  at  their  point  of  intersection  is  perpendicular  to 
their  plane. 


238  SOLID   GEOMETRY.  — BOOK  VI. 

401.  Cor.  I.  Since  E  is  any  point  in  plane  BS,  it  fol- 
lows that 

If  a  plane  is  perpejidicular  to  a  straight  line  at  its  middle 
point,  any  point  in  the  plane  is  equally  distant  from  the  ex- 
tremities of  the  liyie. 

Prop.  IV.     Theorem. 

402.  All  the  perpeiidiculars  to  a  straight  line  at  a  given 
point  lie  in  a  plane  perpendicular  to  the  line. 


Given  AC,  AD,  and  AE  any  three  Js  to  line  AB  at  A. 

To  Prove  that  they  lie  in  a  plane  ±  to  AB. 

Proof.     Let  MN  be  the  plane  determined  by  ^C  and  AD. 
Then,  plane  MNis  ±AB. 

[A  str.  line  ±  to  each  of  two  str.  lines  at  their  point  of  intersection 
is  ±  to  their  plane.]  (§  400) 

Let  the  plane  determined  by  AB  and  AE  intersect  MN 
in  line  AE' ;  then,  AB  _L  AE'. 

[A  ±  to  a  plane  is  ±  to  eveiy  str.  line  drawn  in  the  plane  through 
its  foot.]  (§398) 

But  in  plane  ABE,  only  one  _L  can  be  drawn  to  AB  at  A. 

[At  a  given  point  in  a  str.  line,  but  one  ±  to  the  line  can  be  drawn.] 

(§25) 
Then,  AE'  coincides  with  AE,  and  AE  lies  in  plane  MN. 
.   But  AC,  AD,  and  AE  are  any  three  Js  to  AB  at  A. 
Therefore,  all  the  Js  to  AB  at  A  lie  in  a  plane  J_  AB. 

403.   Cor.  I.     Tlirough  a  given  point  in  a  straight  line,  a 
plane  can  he  drawn  perpendicular  to  the  line,  and  hut  one. 


LINES  AND  PLANES  IN   SPACE. 


239 


404.  Cor.  II  Through  a  given  point  without  a  straight 
line,  a  plane  can  he  drawn  perpendicular  to  the  line,  and  hut 
one. 

Given  point  C  without  line  AB. 

To  Prove  that  a  plane  can  be  drawn 
through  G  ±  AB,  and  but  one. 

Proof.  Draw  line  CB  _L  AB,  and 
let  BD  be  any  other  J_  to  AB  at  B. 

Then,  the  plane  determined  by  BO 
and  BD  will  be  a  plane  drawn  through  G  _L  AB. 

[A  str.  line  _L  to  each  of  two  str.  lines  at  their  point  of  intersection 
is  ±  to  their  plane.]  (§  400) 

Again,  every  plane  through  G  1.  AB  must  intersect  the 
plane  determined  by  AB  and  J5(7  in  a  line  from  G  J_  AB. 

[A  ±  to  a  plane  is  J.  to  every  str.  line  drawn  in  the  plane  through 
its  foot.]  (§  398) 

But  only  one  ±  can  be  drawn  from  G  to  AB. 

[From  a  given  point  without  a  str.  line,  hut  one  ±  can  be  drawn  to 
the  line.  ]  (§  45) 

Then,  every  plane  through  G  ±  AB  must  contain  BG, 
and  be  _L  to  AB  at  B. 

But  only  one  plane  can  be  drawn  through  B  A.  AB. 

[Through  a  given  point  in  a  str.  line,  but  one  plane  can  be  drawn  _L 
to  the  line.]  (§403) 

Hence,  but  one  plane  can  be  drawn  through  G  X  AB. 

405.  Cor.  III.     (Converse  of  §  401.)     Any  point  equally 
distant  from  the  extremities  of  a  straight  line  lies  in  a  plane 
perpendicidar   to  the  line  at  its  mid- 
dle point. 

Given  plane  MN  ±  to  line  ^B  at 
its  middle  point  G,  and  point  D 
equally   distant  from  A  and  B. 

To  Prove  that  D  lies  in  MN. 

(By  §  43,  GD±AB;  then  use  §  402.) 


240 


SOLID   GEOMETRY.  — BOOK   VI. 


Note.     It  follows  from  §§401  and  405  that 

The  locus  (§  141)  of  points  in  space  equally  distant  from  the  ex- 
tremities of  a  straight  line  is  a  plane  perpendicular  to  the  line  at  its 
middle  point. 

Prop.  V.     Theorem. 

406.  If  from  a  point  in  a  perpendicular  to  a  plane,  oblique 
lines  be  drawn  to  the  plane, 

I.    Two  oblique  lines  cutting  off  equal  distances  from  the 
foot  of  the  perpendicular  are  equal. 

II.  Of  two  oblique  lines  cutting  off  unequal  distances  from 
the  foot  of  the  perpendicular,  the  more  remote  is  the  greater. 


I.   Given  line  AB  _L  to  plane  MN  at  B,  and  AC  and  AD 

oblique  lines  meeting  MN  at  equal  distances  from  B. 

To  Prove  AC  =  AD. 

Proof.     Draw  lines  BC  and  BD. 

In  A  ABC  and  ABD,     AB  =  AB. 

Also,  ZABC  =  ZABD. 

[A  ±  to  a  plane  is  ±  to  every  str.  line  drawn  in  the  plane  through 
its  foot.]  (§  398) 

And  by  hyp.,  BC  =  BD. 

.'.  A  ABC  =  A  ABD. 

[Two  A  are  equal  when  two  sides  and  the  included  Z  of  one  are 
equal  respectively  to  two  sides  and  the  included  Z  of  the  other.]    (§  63) 

.-.  AC  =  AD. 

[In  equal  figures,  the  homologous  parts  are  equal.]  (§  66) 


LINES  AND   PLANES  IN  SPACE.  241 

.  II.  Given  line  AB  ±  to  plane  J/iVat  B,  and  AC  and  AE 
oblique  lines  from  A  to  MNj  AE  meeting  MM  at  a  greater 
distance  from  B  than  AO. 

To  Prove  AE  >  AC. 

Proof.     Draw  lines  BC  and  BE. 

On  i?^  take  BF=  BC,  and  draw  line  AF. 

Since  ^F  and  AC  meet  JIO^  at  equal  distances  from  B, 
AF  =  AC. 

[If  from  a  point  in  a  ±  to  a  plane,  oblique  lines  be  drawn  to  the 
plane,  two  oblique  lines  cutting  off  equal  distances  from  the  foot  of 
the  ±  are  equal.]  (§406,1) 

But,  AB  ±  BE. 

[A  J_  to  a  plane  is  ±  to  every  str.  line  drawn  in  the  plane  through 
its  foot.]  (§398) 

.-.  AE  >  AF. 

[If  oblique  lines  be  drawn  from  a  point  to  a  str.  line,  of  two  oblique 
lines  cutting  off  unequal  distances  from  the  foot  of  the  ±  from  the 
point  to  the  line,  the  more  remote  is  the  greater.]  (§  49,  II) 

.-.  AE  >  AC. 

Prop.  VI.     Theorem. 

407.  (Converse  of  Prop.  V.)  If  from  a  point  in  a  perpen- 
dicular to  a  plane,  oblique  lines  he  drawn  to  the  plane, 

I.    Two  equal  oblique  lines  cut  off  equal  distances  from  the 
foot  of  the  perpendicular. 

II.  Of  two  unequal  oblique  lines,  the  greater  cuts  off  the 
greater  distance  from  the  foot  of  the  perpendicular. 

I.  Given  line  AB  ±  to  plane  MN  at  B,  AC  and  AD 
equal  oblique  lines  from  A  to  MM,  and  lines  BC  and  BD. 
(Fig.  of  Prop.  V.) 

To  Prove  BC  =  BD. 

(Prove  A  ABC  and  ABD  equal.) 

II.  Given  line  AB  ±  to  plane  MN  at  B,  and  AC  and  AE 
oblique  lines  from  A  to  MN,  AE  being  >  AC',  also,  lines 
BC  and  BE. 


242 


SOLID   GEOMETRY.  — BOOK  VI. 


To  Prove  BE  >  BC. 

(The  proof  is  left  to  the  pupil.) 


Prop.  VII.     Theorem. 

408.  If  through  the  foot  of  a  perpendicular  to  a  plane  a 
line  be  drawn  at  right  angles  to  ayiy  line  in  the  plane,  the 
line  drawn  from  its  intersection  with  this  line  to  any  point 
in  the  perpendicular  will  he  perpendicular  to  the  line  in  the 
plane. 


Given  line  AB  ±  to  plane  MN  at  A,  line  AE  _L  to  any 
line  CD  in  MN,  and  line  BE  from  E  to  any  point  B  in  AB. 

To  Prove  BE  ±  CD. 

Proof.     On  CD  take  EC  =  ED. 

Draw  lines  AC,  AD,  BC,  and  BD. 
.:  AC=AD. 

[If  a  ±  be  erected  at  the  middle  point  of  a  str.  line,  any  point  in 
the  _L  is  equally  distant  from  the  extremities  of  the  line.]         (§  41,  I) 

.-.  BC=BD. 

[If  from  a  point  in  a  ±  to  a  plane,  oblique  lines  be  drawn  to  the 
plane,  two  oblique  lines  cutting  off  equal  distances  from  the  foot  of 
the  ±  are  equal.]  (§  406,  I) 

Then  since  each  of  the  points  B  and  E  is  equally  distant 
from  G  and  D, 

BE  ±  CD. 

[Two  points,  each  equally  distant  from  the  extremities  of  a  str. 
line,  determine  a  ±  at  its  middle  point.]  (§  43) 

409.  Cor.  I.  From  a  given  point  without  a  plane,  one  per- 
pendicular to  the  plane  can  he  drawn,  and  hut  one. 


LINES  AND  PLANES  IN  SPACE. 


243 


Given  point  A  without  plane  MIT. 

To  Prove  that  a  ±  can  be  drawn 
from  A  to  MN,  and  but  one. 

Proof.  Let  DE  be  any  line  in 
plane  MN;  draw  line  AF  _L  DE, 
line  BF  in  plane  MN  ±  DE,  line 
AB  ±  BF,  and  line  BE. 

Now  EF  is  J_  to  the  plane  determined  by  AF  and  BF. 

[A  str.  line  ±  to  each  of  two  str.  lines  at  their  point  of  intersection 
is  ±  to  their  plane.]  .  (§  400) 

Then  since  BF  is  drawn  through  the  foot  of  EF,  ±  to  line 
AB  in  plane  ABE,  we  have  BE  _L  AB. 

[If  through  the  foot  of  a  ±  to  a  plane  a  line  be  drawn  at  rt.  A  to 
any  line  in  the  plane,  the  line  drawn  from  its  intersection  with  this 
line  to  any  point  in  the  _L  will  be  ±  to  the  line  in  the  plane.]     (§  408) 

Then  AB,  being  ±  to  BE  and  BF,  is  _L  to  MN. 

[A  str.  line  ±  to  each  of  two  str.  lines  at  their  point  of  intersection 
is  ±  to  their  plane.]  (§  400) 

If  possible,  let  AC  be  another  ±  from  A  to  MN;  then 
A  ABC  will  have  two  rt.  A. 

[A  ±  to  a  plane  is  ±  to  every  str.  line  drawn  in  the  plane  through 
its  foot.]  (§398) 

But  this  is  impossible. 

Hence,  but  one  _L  can  be  drawn  from  A  to  MN. 

410.  Cor.  II.  TJie  perpendicular  is  the  shortest  line  that 
can  he  clraivn  from  a  point  to  a  plane. 

Given  AB  the  J_  from  point  A  to  plane  MN,  and  AC  any 
other  str.  line  from  A  to  MJSf.     (Fig.  of  §  409.) 
To  Prove  AB  <  AC. 

Proof.     Draw  line  BC ;  then,  AB  ±  BC 

[A  ±  to  a  plane  is  ±  to  every  str.  line  drawn  in  the  plane  through 
its  foot.]  (§  398) 

.-.  AB<AC. 

[The  ±  is  the  shortest  line  that  can  be  drawn  from  a  point  to  a  str. 
line.]  (§  46) 


244 


SOLID   GEOMETRY.— BOOK  VI. 


Note.     The  distance  of  a  point  from  a  plane  signifies  the  length  of 
the  perpendicular  from  the  point  to  the  plane. 

Prop.  VIII.     Theorem. 

411.   If  two   straight   lines   are  parallel,  a  plane   drawn 
through  one  of  them,  not  coinciding  with  the  plane  of  the 
is  parallel  to  the  other. 

A] \B 

\m  i 


Given  line  AB  II  to  line  CD,  and  plane  MN  drawn  through 
CD,  not  coinciding  with  the  plane  of  the  lis. 

To  Prove  AB  II  MN. 

Proof.  The  lis  AB  and  CD  lie  in  a  plane  which  intersects 
JfiV^inline  CD. 

Hence,  if  AB  meets  MN,  it  must  be  at  some  point  of  CD. 

But  since  AB  is  II  CD,  it  cannot  meet  CD  (§  52). 

Then  AB  and  MN  cannot  meet,  and  are  I!  (§  397). 

Prop.  IX.     Theorem. 

412.  If  a  straight  line  is  parallel  to  a  plane,  the  intersec- 
tion of  the  plane  with  any  plane  drawn  through  the  lirie  is 
parallel  to  the  line. 


A 

M 

B 

/ 

/  G 

/ 

J 

]S 

Given  line  AB  II  to  plane  MN;  and  line  CD  the  intersec- 
tion of  MN  with  any  plane  AD  drawn  through  AB. 
To  Prove  AB  li  CD. 

{AB  and  CD  lie  in  the  same  plane,  and  cannot  meet.) 


LINES  AND  PLANES  IN  SPACE. 


245 


413.  Cor.  If  a  line  and  a  plane  are  parallel,  a  parallel  to 
the  line  through  any  point  of  the  plane  lies  in  the  plane. 

Given  line  AB  II  to  plane  MN;  and  line  CI)  through  any 
point  C  of  MN  II  to  AB.     (Fig.  of  Prop.  IX.) 

To  Prove  that  CD  lies  in  MJS^. 

Proof.  The  plane  determined  by  line  AB  and  point  G 
intersects  MN  in  a  line  II  to  AB. 

[If  a  str.  line  is  ||  to  a  plane,  the  intersection  of  the  plane  with  any- 
plane  drawn  through  the  line  is  ||  to  the  line.]  (§  412) 

But  through  C,  only  one  II  can  be  drawn  to  AB. 
[But  one  str.  line  can  be  drawn  through  a  given  point  |1  to  a  given 
str.  line.]  (§  53) 

Whence,  CD  lies  in  MN. 


Prop.  X.     Theorem. 

414.   If  two  parallel  planes  are  cut  by  a  third  plane,  the 
intersections  are  par'ollel. 


Given  II  planes  MN  and  PQ  cut  by  plane  AD  in  lines  AB 
and  CD,  respectively. 
To  Prove  AB  II  CD. 

{AB  and  CD  lie  in  the  same  plane,  and  cannot  meet.) 

415.  Cor.  Parallel  lines  included  between  parallel  planes 
are  eqtial. 

Given  AC  and  BD  II  lines  included  between  II  planes  MN 
and  PQ.     (Fig.  of  Prop.  X.) 

(Prove  AC  =BD  by  %^  414  and  107.) 


246 


SOLID   GEOMETRY. —BOOK  VI. 


Prop.  XI.     Theorem. 

416.    Through  any  given  straight  line,  a  plane  can  he  drawn 
parallel  to  any  other  straight  line. 

A B 


Given  lines  AB  and  CD. 

To  Prove  that  a  plane  can  be  drawn  through  CD  II  AB. 

(Draw  line  GE  II  AB;  then  use  §  411.) 

Note.  If  AB  is  ||  (7Z>,  an  indefinitely  great  number  of  planes  can 
be  drawn  through  CD  \\  AB  (§411)  ;  otherwise,  bilt  one  such  plane 
can  be  drawn,  for  every  plane  drawn  through  CD  \\  AB  must  contain 
CE  (§  413),  and  but  one  plane  can  be  drawn  through  CD  and  CE. 


Prop.  XII.     Theorem. 

417.    TJirough  a  given  p)oint  a  plane  can  he  drawn  parallel 
to  any  two  straight  lines  in  space. 

B C 


Given  poiut  A  and  lines  BC  and  DE. 
To  Prove  that  a  plane  can  be  drawn  through  A  II  to  BC 
and  DE. 

(The  proof  is  left  to  the  pupil ;  see  §  411.) 

Note.  If  BC  and  DE  are  ||,  an  indefinitely  great  number  of  planes 
can  be  drawn  through  A  \\  to  BC  and  DE  (§  411)  ;  otherwise,  but  one 
such  plane  can  be  drawn. 


LINES  AND   PLANES  IN  SPACE. 


247 


Prop.  XIII.     Theorem. 
418.    Two  perpendiculars  to  the  same  plane  are  parallel. 
AK  G 


'  Given  lines  AB  and  CD  ±  to  plane  MN  at  B  and  D,  respec- 
tively. 

To  Prove  AB  II  CD. 

Proof.     Let  A  be  any  point  of  AB,  and  draw  line  AD. 

Also,   draw  line  BD,  and  line  DF  in  plane  MN  _L  BD. 

.'.  CD±DF. 
[A  ±  to  a  plane  is  JL  to  every  str.  line  drawn  in  the  plane  through 
its  foot.]  (§  398) 

Also,  AD  ±  DF. 

[If  through  the  foot  of  a  ±  to  a  plane  a  line  be  drawn  at  rt.  A  to 
any  line  in  the  plane,  the  line  drawn  from  its  intersection  with  this 
line  to  any  point  in  the  ±  will  be  ±  to  the  line  in  the  plane.]     (§  408) 

Then,  CD,  AD,  and  BD,  being  ±  to  DF  at  D,  lie  in  the 
same  plane. 

[All  the  Js  to  a  str.  line  at  a  given  point  lie  in  a  plane  ±  to  the 
line.]  (§  402) 

Then,  since  points  A  and  B  lie  in  the  plane  of  the  lines 
AD,  BD,  and  CD,  AB  lies  in  this  plane. 

[A  plane  is  a  surface  such  that  the  str.  line  joining  any  two  of  its 
points  lies  entirely  in  the  surface.]  (§9) 

That  is,  AB  and  CD  lie  in  the  same  plane. 
Again,  AB  and  CD  are  ±  BD. 

[A  ±  to  a  plane  is  ±  to  every  str.  line  drawn  in  the  plane  through 
its  foot.] 


.-.  AB  II  CD. 

[Two  Js  to  the  same  str.  line  are  ||.] 


(§54) 


248 


SOLID   GEOMETRY.  — BOOK   VI. 


419.  Cor.  I.     If  one  of  two  parallel  lines  is  perpendicular 
to  a  jjla7ie,  the  other  is  also  perpendicular  to  the  pjlane. 

Given  lines  AB  and   CD   II,  and  a         c 

AB  A.  to  plane  MN. 

To  Prove  CD  _L  MK 

Proof.     A  _L  from  C  to  MN  will       /     b 
be  II  AB.  j^ 

[Two  Js  to  the  same  plane  are  jj.]  (§418) 

But  through  C,  only  one  II  can  be  drawn  to  AB. 

[But  one  str.  line  can  be  drawn  through  a  given  point  11  to  a  given 
str.  line.]  (§  53) 

.-.  CD1.MN. 

420.  Cor.  II.     If  each  of  tiuo  straight  lines  is  parallel  to  a 
third  straight  line,  they  are  parallel  to 
each  other. 

Given  lines  AB  and  CD  II  line  EF. 
To  Prove  AB  II  CD. 
(Draw  plane  MN 1.  EF,  and  prove 
AB  II  CD  by  §§  418  and  419.) 


Prop.  XIV.     Theorem. 
421.    Two  planes  perpendicular  to  the  same  straight  line 


are  parallel. 


My 


N 


Given  planes  MN  and  PQ  ±  to  line  AB. 
To  Prove  MN  II  PQ. 

(Prove  as  in  §  54 ;  by  §  404,  but  one  plane  can  be  drawn 
through  a  given  point  ±  to  a  given  str.  line.) 


LINES  AND   PLANES  IN  SPACE.  249 


Prop.  XV.     Theorem. 

422.    If  each  of  tivo  hitersecting  lines  is  parallel  to  a  plane, 
their  plane  is  parallel  to  the  given  plane. 


Given  lines  AB  and  AC,  in  plane  MN,  II  to  plane  PQ. 

To  Prove    .  MN  II  PQ. 

Proof.  Draw  line  AD  J_  PQ,  and  lines  DE  and  DF  II  to 
AB  and  AC,  respectively;  then  DE  and  DF  lie  in  plane  PQ. 

[If  a  line  and  a  plane  are  II,  a  ||  to  the  line  through  any  point  of  the 
plane  lies  in  the  plane.]  (§413) 

Whence,  AD  is  _L  to  DE  and  DF. 

[A  ±  to  a  plane  is  ±  to  every  str.  line  drawn  in  the  plane  through 
its  foot.]  (§398) 

Therefore,  AD  is  ±  to  AB  and  AC. 

[A  str.  line  ±  to  one  of  two  ||s  is  ±  to  the  other.]  (§  56) 

.-.  ADA.MN. 
[A  str.  line  ±  to  each  of  two  str.  lines  at  their  point  of  intersection 
is  ±  to  their  plane.]  (§  400) 

.-.  MNWPQ. 

[Two  planes  ±  to  the  same  str,  line  are  ||.]  (§  421) 

EXERCISES. 

1.  What  is  the  locus  (§  141)  of  the  perpendiculars  to  a  given 
straight  line  at  a  given  point  ? 

2.  What  is  the  locus  of  points  in  space  equally  distant  from  the 
circumference  of  a  given  circle  ? 

3.  A  line  parallel  to  a  plane  is  everywhere  equally  distant  from  it. 
(Fig.  of  Prop.  IX.     Draw  lines  AC  and  BD±MN.    To  prove 

AC=BD.) 


250  SOLID  GEOMETRY.  — BOOK   VI. 


Prop.  XVI.     Theorem. 

423.   A  straight  line  perpendicular  to  one  of  two  parallel 
planes  is  perpendicular  to  the  other  also. 


Given  MN  and  PQ  II  planes,  and  line  AD  ±  PQ. 

To  Prove  AD  1.  MN. 

Proof.  Pass  two  planes  through  AD,  intersecting  MN 
in  lines  AB  and  AC,  and  PQ  in  lines  DE  and  DF, 
respectively. 

Then,  AB  II  DE,  and  AC  11  DF. 

[If  two  II  planes  are  cut  by  a  third  plane,  the  intersections  are  ||.] 

(§414) 

But  AD  is  ±  to  DE  and  DF. 

[A  ±  to  a  plane  is  _L  to  every  str.  line  drawn  in  the  plane  through 
its  foot.]  (§  398) 

Whence,  AD  is  _L  to  AB  and  AC. 

[A  str.  line  ±  to  one  of  two  ||s  is  ±  to  the  other.]  (§  56) 

.'.  AD1.MN. 

[A  str.  line  ±  to  each  of  two  str.  lines  at  their  point  of  intersection 
Is  J_  to  their  plane.]  (§  400) 

424.  Cor.  I.  Two  parallel  planes  are  everywhere  equally 
distant.     (Note,  p.  244.) 

Given  MN  and  PQ  II  planes.     (Fig.  of  Prop.  XVI.) 

To  Prove  MN  and  PQ  everywhere  equally  distant. 

Proof.     All  lines  which  are  ±  to  both  planes  are  II. 

[Two  Js  to  the  same  plane  are  ||.]  (§  418) 

Therefore,  these  lines  are  all  equal. 

[II  lines  included  between  II  planes  are  equal.]  (§  415) 


/ 

A          / 

p 

N 

B            / 

-^Q 

LINES   AND   PLANES   IN  SPACE.  251 

425.  Cor.  II.  Through  a  given  point  a  plane  can  be  drawn 
parallel  to  a  given  plane,  and  but  one. 

Given  point  A  and  plane  PQ. 

To  Prove  that  a  plane  can  be  drawn 
through  A  II  FQ,  and  but  one. 

Proof.     Draw  line  AB  ±  PQ. 

Through  A  pass  plane  MNX  AB. 

Then  J»/iVr  will  be  II  PQ. 

[Two  planes  ±  to  the  same  str.  line  are  ||.]  (§421) 

If  another  plane  could  be  drawn  through  A  II  PQ,  it  would 
be  ±AB. 

[A  str.  line  J_  to  one  of  two  ||  planes  is  ±  to  the  other  also.]  (§  423) 

It  would  then  coincide  with  MN. 

[Through  a  given  point  in  a  str.  line,  but  one  plane  can  he  drawn 
±  to  the  line.]  (§403) 

Then  but  one  plane  can  be  drawn  through  A  II  PQ. 

EXERCISES. 

4.  "What  is  the  locus  of  points  in  space  equally  distant  from  the 
vertices  of  a  given  triangle  ? 

5.  "What  is  the  locus  of  points  in  space  equally  distant  from  a 
given  plane  ? 

6.  "What  is  the  locus  of  points  in  space  equally  distant  from  two 
parallel  planes  ? 

7.  A  line  parallel  to  each  of  two  intersecting  planes 
is  parallel  to  their  intersection. 

(Pass  a  plane  through  AB  II  PB ;  then  use  §  412.) 

8.  If  two  planes  are  parallel  to  a  third  plane,  they  are  parallel  to 
each  other.     (§§423,421.) 

9.  Line  AB  is  perpendicular  to  plane  MN  at  A.  A  line  is  drawn 
from  A  meeting  any  line  CD  of  plane  MN  at  E.  If  line  BE  is  per- 
pendicular to  CD,  prove  AE  perpendicular  to  CD. 

(Fig.  of  Prop.  VIL) 


252  SOLID   GEOMETRY.— BOOK  VI. 


Prop.  XVII.     Theorem. 

426.  If  two  angles  not  in  the  same  plane  have  their  sides 
parallel  and  exteiiding  in  the  same  direction,  they  are  equal, 
and  their  planes  are  parallel. 


M 

/       ^^^^^ 

/ 

/^f^ 

^P  / 

\ 

.V 

A   \ 

\ 

/\A 

-^\    / 

/  2^^— 

^^C'// 

« 

Given  A  BAC  and  B'A'C  in  planes  MN  and  PQ,  respec- 
tively, with  AB  and  AC  II  respectively  to  A'B'  and  A'C,  and 
extending  in  the  same  direction. 

To  Prove     Z  BAC  =  Z  B'A'C,  and  MW  II  PQ. 

Proof.  Lay  off  AB  =  A'B'  and  AC  =  A'C,  and  draw 
lines  AA',  BB',  CC,  BC,  and  B'C. 

Then  since  AB  is  equal  and  II  to  A'B',  ABB' A'  is  a  O. 

[If  two  sides  of  a  quadrilateral  are  equal  and  ||,  the  figure  is  a  O.J 

(§  110) 
Whence,  A  A'  is  equal  and  II  to  BB'. 

[The  opposite  sides  of  a  /Z7  are  equal.]  (§  106,  I) 

Similarly,  ACCA'  is  a  O,  and  AA'  is  equal  and  II  to  CC. 

Then,  BB'  is  equal  and  11  to  CC. 

[If  each  of  two  str.  lines  is  II  to  a  third  str.  line,  they  are  ||  to  each 
other.]  (§  420) 

Whence,  BB'CC  is  a  O,  and  BC=B'C. 
.'.  AABC=AA'B'C. 

[Two  A  are  equal  when  the  three  sides  of  one  are  equal  respectively 
to  the  three  sides  of  the  other.]  (§  69) 

.-.  Z  BAC  =Z  B'A'C. 

[In  equal  figures,  the  homologous  parts  are  equal.  ]  (§  66) 

Again,  lines  AB  and  AC  are  II  to  plane  PQ. 

[If  two  str.  lines  are  jj,  a  plane  drawn  through  one  of  them,  not 
coinciding  with  the  plane  of  the  jjs,  is  ||  to  the  other.]  (§  411) 


LINES  AND   PLANES  IN  SPACE. 


253 


.-.  MN II  pq. 

[If  each  of  two  intersecting  lines  is  ||  to  a  plane,  their  plane  is  ||  to 
the  given  plane.]  (§422) 

Prop.  XVIII.     Theorem. 

427.   If  tico  straight  lines  are  cat  hy  three  parallel  planes, 
the  corresponding  segments  are  j^roportional 


M 

/^ 



' — iK^ 

P     \\ 

W 

w 

/b[^ 

4/ 

H      \ 

A 

r 

yX 

i/ 

b' 

Given  II  planes  MN,  PQ,  and  RS  intersecting  lines  AC 
and  AC  in  points  A,  B,  0,  and  A',  B',  C,  respectively. 

To  Prove  AB^A^^ 

BC     B'C 
Proof.     Draw  line  AC ;  and  through  AC  and  AC  pass  a 
plane  intersecting  PQ  and  RS  in  lines  BD  and  CC,  respec- 
tively. 

.-.  BD  II  CC. 
[If  two  11  planes  are  cut  by  a  third  plane,  the  intersections  are  ||.] 

(§  414) 

"  BC~  DC'  ^  ^ 

[A  II  to  one  side  of  a  A  divides  the  other  two  sides  proportionally.] 

(§244) 

In  like  manner,  AD^A^  ^2) 

DC     B'C  ^  ^ 

From  (1)  and  (2),  ||  =  ||'. 

[Things  which  are  equal  to  the  same  thing,  are  equal  to  each  other.  ] 

(Ax.  1) 


254 


SOLID   GEOMETRY.— BOOK   VI. 


DIEDRAL   ANGLES. 
DEFINITIONS. 

428.  A  diedral  angle  is  the  amouiit  of  divergence  of  two 
planes  which  meet  in  a  straight  line. 

The  line  of  intersection  of  the  planes  is 
called  the  edge  of  the  diedral  angle,  and  the 
planes  are  called  its  faces. 

Thus,  in  the  diedral  angle  between  planes 
BD  and  BF,  BE  is  the  edge,  and  BD  and  BF 
the  faces. 

A  diedral  angle  may  be  designated  by  two  letters  on  its 
edge ;  or,  if  several  diedral  angles  have  a  common  edge,  by 
four  letters,  one  in  each  face  and  two  on  the  edge,  the 
letters  on  the  edge  being  named  between  the  other  two. 

Thus,  we  may  read  the  above  diedral  angle  BE,  or  ABEC. 

Two  diedral  angles  are  said  to  be  adjacent 
when  they  have  the  same  edge,  and  a  common 
face  between  them;  as,  ABEC  and  CBED. 

Two  diedral  angles  are  said  to  be  vertical 
when  the  faces  of  one  are  the  extensions  of 
the  faces  of  the  other. 

429.  A  plane  angle  of  a  diedral  angle  is  the  angle  be- 
tween two  straight  lines  drawn  one  in  each 
face,  perpendicular  to  the  edge  at  the  same 
point. 

Thus,  if  lines  AB  and  J.C  be  drawn  in  faces 
DE  and  DF,  respectively,  of  diedral  angle 
DG,  perpendicular  to  DG  at  A,  Z  BAG  is  a 
plane  angle  of  the  diedral  angle. 

430.  Let  BAG  and  BA^G  (Fig.  of  §  429)  be  plane  A  of 
diedral  Z  DG]  then,  AB  II  A'B^  and  AG  II  A'G.  (§  54) 

.-.  Z  BAG  =  Z  B'A'G.  (§  426) 

That  is,  all  plane  angles  of  a  diedral  angle  are  equal. 


DIEDRAL  ANGLES. 


255 


431.  A  plane  perpendicular  to  the  edge  of  a  diedral  angle 
intersects  the  faces  in  lines  perpendicular  to  the  edge 
(§  398) ;  hence,  a  plane  perpendicular  to  the  edge  of  a  diedral 
angle  intersects  the  fa.ces  in  lines  which  include  the  plane  angle 
of  the  diedral  angle  (§  429). 

432.  Two  diedral  angles  are  equal  when  their  faces  may 
be  made  to  coincide. 


Prop.  XIX.     Theorem. 

433.    Two  diedral  angles  are  equal  if  their  plane  angles  are 
equal. 

c 


E' 


Given  ABC  and  A'B'C  plane  A  of  diedral  ABB  and 
B'D',  respectively,  and  Z.ABC  =  ZA'B'C'. 

To  Prove       diedral  ZBD  =  diedral  ZB'D'. 

Proof.  Apply  diedral  Z  B'D'  to  BD  in  such  a  way  that 
A'B'  shall  coincide  with  AB,  and  B'C  with  BC. 

Now  BD  and  B'D'  are  ±  to  the  planes  of  A  ABC  and 
A'B'C,  respectively.  (§  400) 

Whence,  B'D'  will  coincide  with  BD.  (§  399) 

Then,  A'D'  will  coincide  with  AD,  and  CD'  with  CD. 

(§  395,  III) 

Hence,  B'D'  and  BD  are  equal.  (§  432) 

434.  Cor.  I.  (Converse  of  Prop.  XIX.)  If  two  diedral 
angles  are  equal,  their  plane  angles  are  equal.  (Fig.  of 
Prop.  XIX.) 

(Apply  ^'Z)'  to  BD  so  that  face  A'D'  shall  coincide  with 
AD,  and  CD'  with  CD,  point  B'  falling  at  B.) 


256 


SOLID   GEOMETRY. —BOOK   VI. 


435.  Cor.  II.  If  two  planes  intersect,  the  vertical  diedral 
angles  are  equal. 

Eor  their  plane  angles  are  equal.  (§  40) 

436.  Defs.  If  a  plane  meets  another  plane  in  such  a  way 
as  to  make  the  adjacent  diedral  angles  equal,  each  is  called  a 
right  diedral  angle,  and  the  planes  are  said  to  be  perpendicu- 
lar to  each  other. 

Thus,  if  plane  PQ  be  drawn  meeting 
plane  MN  in  such  a  way  as  to  make 
diedral  A  PRQM  and  PRQN  equal, 
each  of  these  is  a  right  diedral  Z,  and 
MN  and  PQ  are  ±  to  each  other. 


Prop.  XX.     Theorem. 

437.    Through  a  given  line  m  a  plane,   a  plane  can  he 
drawn  perpendicular  to  the  given  plane,  and  hut  one. 


(Prove  as  in  §  25.) 


Prop.  XXI.     Theorem. 

438.  If  two  planes  are  perpendicular  to  each  other,  a 
straight  line  drawn  in  one  of  them  perpendicular  to  their 
intersection  is  perpendicular  to  the  other. 


M^ 

A 

/ 

/ 

P 

n 

rZ 

/           / 

Z 

/B        /. 

N 


DIEDRAL   ANGLES.  257 

Given  planes  PQ  and  MN  ±,  intersecting  in  line  QR,  and 
line  AB  in  plane  PQ  ±  QR. 

To  Prove  AB  _L  MN. 

Proof.     Draw  line  C'BC  in  plane  MN  1.  QR. 

Then,  ABG  and  ^5C'  are  plane  A  of  diedral  zi  PRQN 
and  PRQM,  respectively.  (§  429) 

Now,  if  two  planes  are  JL  to  each  other,  the  adj.  diedral 
A  are  equal  (§  436). 

That  is,    diedral  A  PRQN=  diedral  A  PRQM. 

.-.  A  ABC  =  A  ABC.  (§434) 

Whence,  A  ABC  is  a  rt.  A.  (§  24) 

Then  ^^,  being  ±  to  56'  and  J5Q  at  B,  is  ±  Jf^.     (§  400) 

439.  Cor.  I.  If  two  planes  are  perpendicular  to  each  other, 
a  perpendicular  to  one  of  them  at  any  point  of  their  intersec- 
tion lies  in  the  other. 

Given  planes  PQ  and  MN  ±,  intersecting  in  line  QR, 
and  line  AB  drawn  from  any  point  B  oi  QR  ±  MN. 
(Fig.  of  Prop.  XXI.) 

To  Prove  that  AB  lies  in  PQ. 

Proof.  If  a  line  be  drawn  in  PQ  from  point  B  ±  QR,  it 
will  be  ±  MN.  (§  438) 

But  from  point  B  but  one  ±  can  be  drawn  to  MN.     (§  399) 
Therefore,  AB  lies  in  PQ. 

440.  Cor.  II.  If  two  planes  are  perpendicidar  to  each 
other,  a  perpendicular  to  one  of  them  from  any  point  of  the 
other  lies  in  the  other. 

Given  planes  PQ  and  MN  ±,  intersecting  in  line  QR,  and 
line  AB  drawn  from  any  point  A  of  PQ  J_  MN.  (Fig.  of 
Prop.  XXI.) 

To  Prove  that  AB  lies  in  PQ. 
(The  proof  is  left  to  the  pupil.) 


258 


SOLID   GEOMETRY.  — BOOK   VI. 


Prop.  XXII.     Theorem. 

441.   If  a  straight  line  is  perpendicular  to  a  pla^ie,  every 
plane  draivn  through  the  line  is  perpendicular  to  the  plane. 


Given  line  AB  ±  plane  MN,  and  PQ  any  plane  drawn 
through  AB. 

To  Prove  PQ  ±  MN. 

Proof.     Let  line  QR  be  the  intersection  of  PQ  and  MN, 
and  draw  line  C'BC  in  plane  MN  ±  QR. 

We  have  AB  ±  BQ.  (§  398) 

Then,  A  ABC  and  ABO  are  plane  A  of  diedral  A  PRQN 

and  PRQM,  respectively.  (§  429) 

But  A  ABC  and  ABC  are  rt.  A.  (§  398) 

.-.  Z  ABC  =  A  ABC.  (§  26) 

.-.  diedral  A  PRQN=diedr2i\  A  PRQM.      (§  433) 

.-.  PQ±MN  (§436) 

Prop.  XXIII.     Theorem. 

442.   A  plane  perpendicular  to  each   of  two   intersecting 
planes  is  perpendicular  to  their  intersection. 


DIEDRAL  ANGLES. 


259 


Given  planes  PQ  and  RS  _L  to  plane  MJST,  and  intersect- 
ing in  line  AB. 

To  Prove  AB  ±  MN. 

(By  §  439,  a  ±  to  JfiV^at  B  lies  in  both  PQ  and  RS.) 

Prop.  XXIV.     Theorem. 

443.   Every  point  in  the  bisecting  plane  of  a  diedral  angle 
is  equally  distant  from  its  faces. 


Given  P  any  point  in  bisecting  plane  BE  of  diedral 
Z  ABDC,  and  lines  PM  and  PJSf  ±  to  AD  and  OZ),  .respec- 
tively. 

To  Prove  PM=PN. 

Proof.  Let  the  plane  determined  by  PM  and  PN  inter- 
sect planes  AD,  BE,  and  CD  in  lines  FM,  FP,  and  FN, 
respectively. 

Plane  PMFNis  _L  to  planes  AD  and  CD.  (§  441) 

Then,  plane  PMFN  is  ±  BD.  (§  442) 

Whence,  Pi^Jf  and  PFN  are  plane  Zs  of  diedral  A  ABDE 

and  CBDE,  respectively.  (§  431) 

.-.  ZPFM=^ZPFN.  (§434) 

In  A  PP3/  and  PEN,  PF  =  PF. 

And,  APFM=Z.PFN. 

Also,     zl  PJltfP  and  PNE  are  rt.  A  (§  398) 

.-.  APFM=APFK  (§  70) 

.-.  PM=PN.  (?) 


260  SOLID   GEOMETRY.— BOOK  VI. 

444.  Cor.  I.  (Converse  of  Prop.  XXIV.)  A7iy  point 
ivhich  is  ivithin  a  diedi'ol  angle,  and  equally  distant  from  its 
faces,  lies  in  the  bisecting  plane  of  the  diedral  angle. 

Given  point  P  within  diedral  Z.ABDC,  equally  distant 
from  AD  and  CD,  and  plane  BE  determined  by  BD  and  P. 
(Fig.  of  Prop.  XXIV.) 

To  Prove  that  BE  bisects  diedral  Z  ABDC. 

(Prove  A  PFM  and  PEN  equal ;  then  Z  PFM  =  Z  PEN, 
and  the  theorem  follows  by  §  433.) 

445.  Cor.  II.     It  follows  from  §§  443  and  444  that 

The  locus  of  points  in  space  equally  distant  from  the  faces 
of  a  diedral  angle  is  the  plane  bisecting  the  diedral  angle. 

Pkop.  XXV.     Theorem. 

446.  Through  a  given  straight  li?ie  ivithout  a  plane,  a  plane 
can  be  drawn  perpendicular  to  the  given  plane,  and  but  one. 


Given  line  AB  without  plane  MN. 

To  Prove  that  a  x^lane  can  be  drawn  through  AB  ±  MN, 
and  but  one. 

Proof.  Draw  line  ACA-MN,  and  let  AD  be  the  plane 
determined  by  AB  arid  AC;  then,  AD1.MN.  (§  441) 

If  more  than  one  plane  could  be  drawn  through  AB  _L  MN, 
their  common  intersection,  AB,  would  be  ±  MN.        (§  442) 

Hence,  but  one  plane  can  be  drawn  through  ABA.MN, 
unless  AB  is  ±  MN. 

Note.  If  Hne  AB  is  ±  MN.,  an  indefinitely  great  number  of  planes 
can  be  drawn  through  AB  1.MN  (§  441). 


DIEDRAL  ANGLES. 


261 


447.  Def s.  The  projection  of  a  point  on  a  plane  is  the  foot 
of  the  perpendicular  drawn  from  the  point  to  the  plane. 

The  projection  of  a  line  on  a  plane  is  the  line  which  con- 
tains the  projections  of  all  its  points. 

448.  Cor.  The  iwojection  of  a  straight  line  on  a  plane  is 
a  straight  line. 

Given  line  CD  the  projection  (§  447)  of  str.  line  AB  on 
plane  MN.     (Fig.  of  Prop.  XXV.) 

To  Prove  CD  a  str.  line. 

Proof.     Draw  a  plane  through  ABA-MK 

The  Js  to  MN  from  all  points  of  AB  will  lie  in  this 
plane.  (§  440) 

Therefore,  CD  is  a  str.  line.  (§  396) 

Prop.  XXVI.     Theorem. 

449.  The  angle  between  a  straight  hne  and  its  projection 
on  a  plane  is  the  least  angle  which  it  makes  with  any  line 
drawn  in  the  plane  through  its  foot. 


N 

Given  line  BC  the  projection  of  line  AB  on  plane  MN, 
and  BD  any  other  line  drawn  through  B  in  MN. 
To  Prove  Z  ABC  <  Z  ABD. 

Proof.     Lay  off  BD  =  BC,  and  draw  lines  AC  and  AD. 
In  A  ABC  and  ABD,  AB  =  AB. 
And  by  hyp.,  BC  =  BD. 

Also,  AC<AD.  (§410) 

.-.  ZABC<ZABD.  (§92) 

Note.    Z  ABC  is  called  the  angle  between  line  AB  and  plane  MN. 


262 


SOLID   GEOMETRY.  — BOOK  VI. 


Prop.  XXVII.     Theorem. 

450.  Two  straight  lines,  not  in  the  same  plane,  have  one 
coynmon  j^^^'P^ndicular,  and  but  one;  and  this  line  is  the 
shortest  line  that  can  be  drawn  between  them. 


Given  lines  AB  and  CD,  not  in  the  same  plane. 

To  Prove  that  one  common  ±  to  AB  and  CD  can  be 
drawn,  and  but  one ;  and  that  this  line  is  the  shortest  line 
that  can  be  drawn  between  AB  and  CD. 

Proof.     Through  CD  draw  plane  MN  II  AB.  (§  416) 

Through  AB  draw  plane  AH  1.  MN,  and  produce  their 


intersection  to  meet  CD  at  G. 

Draw  line  AG  in  plane  AH  1.  GH 


(§  446) 
then,  AG  ±  MN. 

(§  438) 

(§  398) 

(§  412) 

(§56) 


.-.  AGXCD. 
Also,  GH  II  AB. 

.'.  AG±AB. 
Then,  AG  is  a  common  J_  to  AB  and  CD. 
If  possible,  let  EKhe  another  common  ±  to  AB  and  CD, 

and  draw  line  EF  II  AB,  and  line  KL  in  plane  AH  J_  GH. 
Then,  EF  lies  in  plane  MN.  (§  413) 

Also,  EK  is  _L  to  ED  and  EF.  (§  56) 

Whence,  EK  is  _L  MN.  (?) 

But  KL  is  also.X  JWiV.  (§  438) 

We  should  then  have  two  Js  from  K  to  MN,  which  is 

impossible.  (§  409) 

Hence,  but  one  common  ±  can  be  drawn  to  AB  and  CD. 
Again,  EK  >  KL.  (§  410) 


DIEDRAL  ANGLES. 


263 


.'.EK>AG.  (§80) 

Hence,  AG  \^  the  shortest  line  between  AB  and  CD. 

Prop.  XXVIII.     Theorem. 

451.    Two  diedral  angles  are  to  each  other  as  their  plane 
angles. 

Case  I.     When  the  plane  angles  are  commensurable. 


i>W~ 


-T-^-.- 

3^ 


Dl^-^: 


Given  ABC  and  A'B'C,  plane  A  of  diedral  A  ABDC  and 
A'B'IJ'C,  respectively,  and  commensurable. 


To  Prove 


ABDC        Z  ABC 


A'B'D'C     ZA'B'C 


Proof.  Let  Z  ABE  be  a  common  measure  of  A  ABC  and 
A'B'C;  and  suppose  it  to  be  contained  4  times  in  ZABC 
and  3  times  in  Z^'5'0'. 

ZABC  ^4  ... 

'*  Z  A'B'C     3  ^  ^ 

Passing  planes  through  edges  BD  and  B'D',  and  the 
several  lines  of  division  of  A  ABC  and  A'B'C,  respectively, 
diedral  ZABDC  will  be  divided  into  4  parts,  and  diedral 
Z  A'B'D'C  into  3  parts,  all  of  which  parts  are  equal.    (§  433) 

ABDC       4 


A'B'D'C     3 

T.         ...       WON     ABDC        ZABC 
Prom  (1)  and  (2),^,^^,  =  ^-^^, 


(2) 


264  SOLID   GEOMETRY.— BOOK  VI. 

Case  II.    Whe7i  the  plane  angles  are  incommensurdble. 


c 


Given  ABC  and  ABC  plane  A  of  diedral  A  ABDC  and 
A'B'D'C,  respectively,  and  incommensurable. 

To  Prove  ABDC  ^  Z  ABC  ^ 

A'B'D'C     AA'B'C 

Proof.  Let  Z  ABC  be  divided  into  any  number  of  equal 
parts,  and  let  one  of  these  parts  be  applied  to  Z  A'B'C  as  a 
unit  of  measure. 

Since  A  ABC  and  A'B'C  are  incommensurable,  a  certain 
number  of  the  parts  will  extend  from  A'B'  to  B'E,  leaving 
a  remainder  ZEB'C  <  one  of  the  equal  parts. 

Pass  a  plane  through  B'D'  and  B'E\  then  since  the  plane 
A  of  diedral  AA'B'D'E  and  ABDC  are  commensurable, 

ABDC  ^Z  ABC  ^        (§451,  Case  I) 
A'B'D'E      ZA'B'E         ^  '  ^ 

Now  let  the  number  of  subdivisions  of  Z  ABC  be  indefi- 
nitely increased. 

Then  the  unit  of  measure  will  be  indefinitely  diminished, 
and  the  remainder  Z  EB'C  will  approach  the  limit  0. 

^,          ABDC      -n  1   ^v.    T    -4.    ABDC 

Then will  approach  the  limit 


and 


A'B'D'E 
ZABC 


will  approach  the  limit 


A'B'D'C 
ZABC 


Z  A' B'E  '■  •■  Z  A'B'C 

By  the  Theorem  of  Limits,  these  limits  are  equal.  (§  188) 

ABDC  ^  ZABC 
' '  A'B'D'C      Z  A'B'C' 


DIEDRAL   ANGLES. 


265 


Note.  It  follows  from  §  451  that  the  plane  angle  may  be  taken  as 
the  measure  of  the  diedral  angle  ;  thus,  if  the  plane  angle  contains  n 
degrees,  the  diedral  angle  may  be  regarded  as  being  of  n  degrees. 


EXERCISES. 

10.  A  straight  line  and  a  plane  perpendicular  to  the  same  straight 
line  are  parallel. 

(Fig.  of  Prop.  IX.     Let  plane  determined  by  ^B  and  AC  inter- 
sect MN  in  CD.) 

11.  If  two  planes  are  parallel,  a  line  parallel  to  one  of  them  through 
any  point  of  the  other  lies  in  the  other. 

(Fig.  of  Prop.  X.     Given  planes  MN  and  PQ  ||,  and  AB  through 
any  point  A  of  MN  II  PQ.     Prove  that  AB  lies  in  MN  by  §  413.) 


12.  If  a  straight  line  is  parallel  to  a  plane, 
any  plane  perpendicular  to  the  line  is  perpendicu- 
lar to  the  plane. 

(Draw  line  CD  ±  QE,  and  prove  it  ±  MN) 


13.   If  two  parallels  meet  a  plane,  they  make 
equal  angles  with  it. 
(Given  AB  \\  CD ;  to  prove  ZABA'  =ZCDC'.) 


14.    If  a  straight  line  intersects  two  parallel  planes,  it  makes  equal 
angles  with  them. 


15.  The  angle  between  perpendiculars  to  the 
faces  of  a  diedral  angle  from  any  point  within 
the  angle  is  the  supplement  of  its  plane  angle. 

(Prove  ZBDC  the  plane  Z  of  diedral  ZPQBS. ) 


16.   If  each  of  two  intersecting  planes  be  cut 
by  two  parallel  planes,  not  parallel  to  their  inter- 
section,  their    intersections    with    the    parallel  ^ 
planes  include  equal  angles. 

(To  prove  ZABC=Z  DEF. ) 


/ 

B 
7 

M/— 

„i?... 

/ 

/ 

/ 

L^ 

/ 

D         E 
/ 

/ 

R               f 

M        / 

<    A' 

//. 

W>/ 

/- 

^    / 

N 

266  SOLID   GEOMETRY.— BOOK  VI. 

POLYBDRAL  ANGLES. 
DEFINITIONS. 

452.  A  polyedral  angle  is  a  figure  composed  of  three  or 
more  triangles,  called /aces,  having  for  their 

bases  the  sides  of  a  polygon,  and  for  their  9 

common  vertex  a  point  without  its  plane ;  //l\ 

asO-ABCD.  /      \\ 

The  common  vertex,  0,  is  called  the  ver-         /     IId  \ 
tex  of  the  polyedral  angle,  and  the  polygon,    "^  vT    /   ^^^^ 
ABCD,  the  base  ;  the  vertical  angles  of  the  ^^^^^ 

triangles,  AOB,  BOC,  etc.,  are  called  the 
face  angles,  and  their  sides,  OA,  OB,  etc.,  the  edges. 

Note.  The  polyedral  angle  is  not  regarded  as  limited  by  the  base  ; 
thns,  the  face  AOB  is  understood  to  mean  the  indefinite  plane  between 
the  edges  OA  and  OB  produced  indefinitely. 

A  triedral  angle  is  a  polyedral  angle  of  three  faces. 
Two  polyedral  angles  are  called  vertical  when  the  edges 
of  one  are  the  prolongations  of  the  edges  of  the  other. 

453.  A  polyedral  angle  is  called  convex  when  its  base  is 
a  convex  polygon  (§  121). 

454.  Two  polyedral  angles  are  equal  when  they  can  be 
applied  to  each  other  so  that  their  faces  shall  coincide. 

455.  Two  polyedral  angles  are  said  to  be  symmetrical 
when    the    face    and    diedral 

angles  of  one  are  equal  respec-  a  y^ 

tively  to  the  homologous  face  //  \  /  \\ 

and  diedral  angles  of  the  other,       /    /     \  /     \     \ 

if  the  equal  parts  occur  in  the  A^—/—--^C      c^^^^Ar--^^' 
reverse  order.  jb  ^b' 

Thus,  if  face  A  AOB,  BOO, 
and  COA  are  equal  respectively  to  face  AA'O'B',  B'O'C, 
and   COA',  and   diedral   A  OA,   OB,  and   OC  to  diedral 
A  OA',  O'B',  and  O'C,  triedral  A  0-ABC  and  O'-A'B'C 
are  symmetrical. 


POLYEDRAL  ANGLES. 


267 


It  is  evident  that,  in  general,  two  symmetrical  polyedral 
angles  cannot  be  placed  so  that  their  faces  shall  coincide. 

Prop.  XXIX.     Theorem. 
456.    Two  vertical  polyedral  angles  are  symmetrical. 

B' 


Given  0-ABC  s^nd  O-A'B'C  (Fig.  1)  vertical  triedral  A. 
To  Prove  0-ABC  Siud  O-A'B'C  symmetrical. 

Proof.  Face  A  AOB,  BOC,  etc.,  are  equal,  respectively, 
to  face  A  A' OB',  B'OC,  etc.  (§  40) 

Again,  diedral  A  OA  and  OA'  are  vertical ;  for  AOB  and 
A' OB'  are  portions  of  the  same  plane,  as  also  are  AOC  and 
A'OC ;  in  like  manner,  diedral  A  OB  and  OB'  are  vertical ; 
etc. 

Then,  diedral  A  OA,  OB,  etc.,  are  equal,  respectively,  to 
diedral  A  OA',  OB',  etc.  (§  435) 

But  the  equal  parts  of  the  triedral  A  occur  in  the  reverse 
order;  as  may  be  seen  by  conceiving  O-A'B'C  moved  II  to 
itself  to  the  right,  and  then  revolved,  as  shown  in  Fig.  2, 
about  an  axis  passing  through  0,  until  face  OA'C  comes 
into  the  same  plane  as  before ;  edge  OB'  being  on  this  side 
of,  instead  of  beyond,  plane  OA'C. 

Hence,  0-ABC  and  O-A'B'C  are  symmetrical  (§  455). 

In  like  manner,  the  theorem  may  be  proved  for  any  two 
polyedral  A. 


268  SOLID   GEOMETRY.  — BOOK   VI. 

Ex.  17.    If  two  parallel  planes  are  cut  by  a  third  plane,  the  al- 
ternate-interior diedral  angles  are  equal. 

(Prove  the  plane  A  of  the  alt. -int.  diedral  A  equal.) 

Prop.  XXX.     Theorem. 

457.    The  sum  of  any  two  face  angles  of  a  triedral  angle  is 
greater  than  the  third. 

Note.     The  theorem  requires  proof  only  in  the  case  where  the  third 
face  angle  is  greater  than  either  of  the  others. 


Given  in  triedral  Z  0-ABC, 

face  ZAOC>  face  Z.AOB  or  face  Z  BOO. 
To  Prove        Z  AOB -\- Z  BOC >  Z  AOC. 

Proof.  In  face  AOC  draw  line  OD  equal  to  OB,  making 
Z  AOD  =  Z  AOB]  and  through  B  and  D  pass  a  plane  cut- 
ting the  faces  of  the  triedral  Z  in  lines  AB,  BG,  and  CA, 
respectively. 

In  AAOB  and  AOD,    OA  =  OA. 

And  by  cons.,  OB  =  OD, 

and  ZAOB  =  ZAOD. 

.-.  AAOB  =  AAOD.  (?) 

.-.  AB  =  AD.  (?) 

Now,  AB-\-BC>AD-^  DO.  (Ax.  4) 

Or,  since  AB  =  AD,      BC  >  DC. 

Then,  in  A  BOG  and  GOD,  OG  =  OG. 


POLYEDRAL  ANGLES.  269 

Also,  OB  =  OD,  and  BC  >  DC. 

.-.  ZB0C>ZC0IJ.  (§  91) 

Adding  Z.AOB  to  the  first  member  of  this  inequality, 
and  its  equ^l  Z  AOD  to  the  second  member,  we  have 
^AOB+ZBOC>AAOD  +  ZCOD. 
.:  ZAOB-hZBOOZAOa 

Prop.  XXXI.     Theorem. 

458.    The  sum  of  the  face  angles  of  any  convex  polyedral 
angle  is  less  than  four  right  angles. 


Given  0-ABCDE  a  convex  polyedral  Z. 

To  Prove  Z  AOB  -f-  Z  BOC  +  etc.  <  4  rt.  A. 

Proof.     Let  ABCIJE  he  the   base   of  the  polyedral  Z. 

Let  0'  be  any  point  within  polygon  ABODE,  and  draw 
lines  O'A,  O'B,  O'O,  O'D,  said  O'E. 

Then,  in  triedral  ZA-EOB,  .- 

Z  OAE  +  Z  OAB  >  Z  O'AE  +  Z  O'AB.      (§  457) 

Also,    Z  OB  A  -h  Z  OBC  >  Z  O'B  A  +  Z  O'BC;  etc. 

Adding  these  inequalities,  we  have  the  sum  of  the  base  A 
of  the  A  whose  common  vertex  is  0  >  the  sum  of  the  base 
A  of  the  A  whose  common  vertex  is  0'. 

But  the  sum  of  all  the  A  of  the  A  whose  common  vertex 
is  0  is  equal  to  the  sum  of  all  the  A  of  the  A  whose  com- 
mon vertex  is  0'.  (§  84) 

Hence,  the  sum  of  the  zi  at  0  is  <  the  sum  of  the  ^  at  0'. 

Then,  the  sum  of  the  Zs  at  0  is  <  4  rt.  A.  (§  35) 


270 


SOLID   GEOMETRY.— BOOK   VI. 


Prop.  XXXII.     Theorem. 

459.  If  two  triedrdl  angles  have  the  face  angles  of  oyie 
equal  respectively  to  the  face  angles  of  the  other,  their  homolo- 
gous diedral  angles  are  equal. 

0'  O' 


Fig.  1 


Given,  in  triedral  A  0-ABC  and  0'-^'J5'(7', 

AAOB  =  AA'0'B',   ZBOC=  Z  B'O'C, 
and  ZCOA  =  ZC'0'A'. 
To  Prove  diedral  ZOA  =  diedral  Z  O'A'. 
Proof.     Lay  off  OA,  OB,  OC,  O'A',  O'B',  and  O'C  all 
equal,  and  draw  lines  AB,  BC,  CA,  A'B',  B'C,  and  C'A'. 

.-.  A  OAB  =  A  O'A'B'.  (§  63) 

.-.  AB  =  A'B'.  (§  66) 

Similarly,        BC  =  B'C  and  CA  =  C'A'. 

.:  A  ABC  =  A  A'B'C.  (§  69) 

.-.  ZEAF=ZE'A'F'.  (?) 

On  OA  and  O'A'  take  AD  =  A'D'. 
Draw  line  DE  in  face  OAB±  OA. 

Since  A  OAB  is  isosceles,  Z  OAB  is  acute,  and  hence  DE 
will  meet  AB ;  let  it  meet  AB  at  E. 

Also,  draw  line  DF  in  face  OACJ-  OA,  meeting  AC  slI  F; 
and  lines  D'E'  and  D'F'  in  faces  O'A'B'  and  0'A'C±0'A', 
meeting  A'B'  and  A'C  at  E'  and  jP',  respectively. 
Draw  lines  EF  and  E'F. 
Then,  in  rt.  A  ADE  and  A' D'E', 
AD  =  ^'Z>'. 


POLYEDRAL  ANGLES. 


271 


And  since  A  OAB  =  A  O'A'B', 

Z  DAE  =  Z  D'A'E'.  (?) 

.:  AADE  =  AA'D'E'.    -  (§89) 

.-.  AE  =  A'E',  and  DE  =  D'E'.  (?) 

Similarly,  AF=  A'F',  and  DF  =  D'F'. 

Then,  in  A  AEF  and  A'E'F', 

AE  =  A'E',  AF=A'F',  and  Z  EAF=  Z  E'A'F'. 

.'.  A  AEF  =  A  A'E'F'.  (?) 

.-.  EF=  E'F'.  (?) 

Then,  in  A  DEF  and  D'E'F', 

DE  =  D'E',  DF=  D'F',  and  EF=  E'F'. 

.:  A  DEF  =  A  D'E'F.  (?) 

.-.  ZEDF=ZE'D'F'.         '  (?) 

But,  EDF  and  E'D'F'  are  the  plane  A  of  diedral  A  OA 

and  0'^',  respectively.  (§  429) 

.-.  diedral  ZOA  =  diedral  Z  0'^'.  (§  433) 

Note.  The  above  proof  holds  for  Fig.  3  as  well  as  for  Fig.  2  ;  in 
Figs.  1  and  2,  the  equal  parts  occur  in  the  same  order,  and  in  Figs.  1 
and  3  in  the  reverse  order. 

460.  Cor.  If  two  triedral  angles  have  the  face  angles  of  one 
equal  respectively  to  the  face  angles  of  the  other, 

1.  They  are  equal  if  the  equal  parts  occur  in  the  same  order. 
For  if  triedral  Z  O'-A'B'C  (Fig.  2)  be  applied  to  0-ABC 

so  that  diedral  A  O'A'  and  OA  coincide,  point  0'  falling 
at  0,  then  since  Z  A'O'C  =  Z  AOC,  and  Z  A'O'B'  =Z  AOB, 
O'B'  will  coincide  with  OB,  and  O'C  with  OC. 

2.  They  are  symmetrical  if  the  equal  parts  occur  in  the 
reverse  order. 

EXERCISES. 

18.  If  ^C  is  the  projection  of  line  AB  upon 
plane  3IN,  and  BD  and  BE  be  drawn  in  the  plane 
making ZCBD^ZCBE,  prove ZABD  =  ZABE. 

(Lay  off  BD  =  BE,  and  draw  lines  AD,  AE. 
CD,  and  CE.     Prove  /^ABD  and  ABE  equal.) 


272 


SOLID   GEOMETRY. —BOOK  VI. 


19.  If  a  plane  be  drawn  through  a  diagonal 
of  a  parallelogram,  the  perpendiculars  to  it  from 
the  extremities  of  the  other  diagonal  are  equal. 

(Given  plane  EF  through  diagonal  ^C  of 
CJABCD;  toipro\eBG=DH.    Prove  rt.  i^BGO^' 
and  Z)ifO  equal. )  q- 

20.  Two  triedral  angles  are  equal  when  a  face  angle  and  the  ad- 
jacent diedral  angles  of  one  are  equal  respectively  to  a  face  angle  and 
the  adjacent  diedral  angles  of  the  other,  and  similarly  placed. 

21.  D  is  any  point  in  perpendicular  AF  from 
A  to  side  BC  of  triangle  ABC.  If  line  DE  be 
drawn  perpendicular  to  the  plane  of  ABC,  and 
line  GH  through  E  parallel  to  BC,  prove  line 
AE  perpendicular  to  GH. 

(Prove  BC±  to  plane  AED  by  §  438.) 

22.  A  is  any  point  in  face  EG  of  diedral  ZDEFG. 
If  ^Cbe  drawn  perpendicular  to  edge  EF,  and  AB 
perpendicular  to  face  DF,  prove  the  plane  deter- 
mined by  ylC  and  BC  perpendicular  to  EF.  (Ex. 
9.) 

23.  From  any  point  E  within  diedral  Z  CABD, 
EF  a-Tid  EG  are  drawn  perpendicular  to  faces  ABC 
and  ABD,  respectively,  and  GH  perpendicular  to 
face  ABC  at  H.     Prove  FH  perpendicular  to  AB. 

(Prove  that  FH  lies  in  the  plane  of  EF  and  EG.) 

24.  The  three  planes  bisecting  the  diedral 
angles  of  a  triedral  angle  meet  in  a  common 
straight  line. 

(Let  planes  GAD  and  QBE  intersect  in  line  A 
OG.     Prove  G  in  plane  OCF  hy  ^  444.) 

25.  Any  point  in  the  plane  passing  through  the  bisector  of  an 
angle,  perpendicular  to  its  plane,  is  equally  distant  from  the  sides 
of  the  angle. 

26.  Any  face  angle  of  a  polyedral  angle  is  less  than  the  sum  of 
the  remaining  face  angles. 

(Divide  the  polyedral  Z  into  triedral  A  by  passing  planes  through 
any  lateral  edge.) 


Book  VII. 

POLYEDRONS. 
DEFINITIONS. 

461.  Apolyedron  is  a  solid  bounded  by  polygons. 

The  bounding  polygons  are  called  the  faces  of  the  polye- 
dron;  their  sides  are  called  the  edges,  and  their  vertices 
the  vertices. 

A  diagonal  of  a  polyedron  is  a  straight  line  joining  any 
two  vertices  not  in  the  same  face. 

462.  The  least  number  of  planes  which  can  form  a  polye- 
dral  angle  is  three. 

Whence,  the  least  number  of  polygons  which  can  bound 
a  polyedron  is  four. 

A  polyedron  of  four  faces  is  called  a  tetraedron;  of  six 
faces,  a  hexaedron;  of  eight  faces,  an  octaedron;  of  twelve 
faces,  a  dodecaedron;  of  twenty  faces,  an  icosaedron. 

463.  A  polyedron  is  called  convex  when  the  section  made 
by  any  plane  is  a  convex  polygon  (§  121). 

All  polyedrons  considered  hereafter  will  be  understood  to 
be  convex. 

464.  The  volume  of  a  solid  is  its  ratio  to  another  solid, 
called  the  7init  of  volume,  adopted  arbitrarily  as  the  unit  of 
measure  (§  180). 

The  usual  unit  of  volume  is  a  cube  (§  474)  whose  edge  is 
some  linear  unit;  for  example,  a  cubic  inch  or  a  cubic  foot. 

465.  Two  solids  are  said  to  be  equivalent  when  their  vol- 
umes are  equal. 

273 


274 


SOLID   GEOMETRY.  — BOOK   VII. 


PRISMS  AND  PARALLBLOPIPBDS. 
DEFINITIONS. 

466.  A  prism  is  a  polyedron,  two  of  whose  faces  are 
equal   polygons  lying  in  parallel   planes, 
having    their   homologous    sides   parallel, 
the  other  faces  being  parallelograms  (§  110). 

The  equal  and  parallel  faces  are  called 
the  bases  of  the  prism,  and  the  other  faces 
the  lateral  faces;  the  edges  which  are  not 
sides  of  the  bases  are  called  the  lateral  edges,  and  the  sum 
of  the  areas  of  the  lateral  faces  the  lateral  area. 

The  altitude  is  the  perpendicular  distance  between  the 
planes  of  the  bases. 

467.  The  following  is  given  for  convenience  of  reference: 
The  bases  of  a  prism  are  equal. 

468.  It  follows  from  the  definition  of  §  466  that  the  lat- 
eral edges  of  a  prism  are  equal  and  parallel.  (§  106,  I) 

469.  A  prism   is   called   triangular,   quadrangular,   etc., 
according  as  its  base  is  a  triangle,  quadrilateral,  etc. 

470.  A  right  prism  is  a  prism  whose  lat- 
eral edges  are  perpendicular  to  its  bases. 

The  lateral  faces  are  rectangles  (§  398). 
An  oblique  prism  is  a  prism  whose  lateral 
edges  are  not  perpendicular  to  its  bases. 


471.  A  regular  prism  is  a  right  prism  whose  base  is  a 
regular  polygon. 

472.  A  truncated  prism  is  a  portion  of  a 
prism  included  between  the  base,  and  a  plane, 
not  parallel  to  the  base,  cutting  all  the  lateral 
edges. 

The  base  of  the  prism  and  the  section  made 
by  the  plane  are  called  the  bases  of  the  trun- 
cated prism. 


PRISMS   AND  PARALLELOPIPEDS. 


275 


473.  A  right  section  of  a  prism  is  a  section  made  by  a 
plane  cutting  all  the  lateral  edges,  and  perpendicular  to  them. 

474.  A  parallelopiped  is  a  prism  whose 
bases  are  parallelograms;  that  is,  all  the 
faces  are  parallelograms. 

A  right  parallelopiped  is  a  parallelopiped 
whose  lateral  edges  are  perpendicular  to 
its  bases. 

A  rectangular  parallelopiped  is  a  right 
parallelopiped  whose  bases  are  rectangles ; 
that  is,  all  the  faces  are  rectangles. 

A  cube  is  a  rectangular  parallelopiped  whose  six  faces  are 
all  squares. 

Prop.  I.     Theorem. 

475.  The  sections  of  a  prism  made  by  two  parallel  planes 
which  cut  all  the  lateral  edges,  are  equal  polygons. 


Given  II  planes  OF  and  C'F'  cutting  all  the  lateral  edges 
of  prism  AB. 

To  Prove  section  CDEFG  =  section  C'D'E'F'G'. 

Proof.     We  have  CD  II  CD',  DE  II  D'E\  etc.  (§  414) 

...  CD=  CD',  DE  =  D'E',  etc.  (§  107) 

Also  Z  CDE  =  Z  CD'E',  Z  DEF=  Z  D'E'F',  etc.  (§  426) 

Then,  polygons  CDEFG  and  C'D'E'FG',  being  mutually 
equilateral  and  mutually  equiangular,  are  equal.         (§  124) 


276 


SOLID   GEOMETRY.— BOOK   VII 


476.   Cor.     The  section  of  a  prism  made  by  a  plane  paral- 
lel to  the  base  is  equal  to-  the  base. 


Prop.  II.     Theorem. 

477.  Two  prisms  are  equal  when  the  faces  including  a  tne- 
dral  angle  of  one  are  equal  respectively  to  the  faces  including 
a  triedral  angle  of  the  other,  and  similarly  placed. 


Given,  in  prisms  AH  and  A'H',  faces  ABCDE,  AG,  and 
AL  equal  respectively  to  faces  A'B'C'D'E',  A'G',  and  A'L'-, 
the  equal  parts  being  similarly  placed. 

To  Prove    prism  AH  =  prism  A'H'. 

Proof.  We  have  AEAB,  EAF,  and  FAB  equal  respec- 
tively to  AE'A'B',  E'A'F',  and  F'A'B'.  (§  66) 
.-.  triedral  Z  A-BEF  =  triedral  Z.  A'-B'E'F'.  (§  460,  1) 

Then,  prism  A'H'  may  be  applied  to  prism  AH  in  such  a 
way  that  vertices  A',  B',  C,  U,  E',  G',  F',  and  L'  shall  fall 
at  A,  B,  C,  D,  E,  G,  F,  and  L,  respectively. 

Now  since  the  lateral  edges  of  the  prisms  are  II,  edge 
OH  will  fall  on  CH,  B'K'  on  DK,  etc.  (§  53) 

And  since  points  6r',  F,  and  V  fall  at  G,  F,  and  L,  respec- 
tively, planes  LH  and  L'H'  coincide.  '  (§  395,  II) 

Then  points  H'  and  K'  fall  at  H  and  K,  respectively. 

Hence,  the  prisms  coincide  throughout,  and  are  equal. 

478.  Cor.  Two  right  prisms  are  equal  when  they  have 
equal  bases  and  equal  altitudes  ;  for  by  inverting  one  of  the 
prisms  if  necessary,  the  equal  faces  will  be  similarly  placed. 


PRISMS  AND  PARALLELOPIPEDS.  277 

479.  Sch.  The  demonstration  of  §  477  applies  without 
change  to  the  case  of  two  truncated  prm/is. 

Prop.  III.     Thi:orem. 

480.  An  oblique  prism  is  equivalent  to  a  right  prison  hav- 
ing for  its  base  a  right  section  of  the  oblique  prism,  and  for 
its  altiticde  a  lateral  edge  of  the  oblique  prism. 


B  C 

Given  FK'  a  right  prism,  having  for  its  base  FK  a  right 
section  of  oblique  prism  AD',  and  its  altitude  FF'  equal  to 
AA',  a  lateral  edge  of  AD'. 

To  Prove  AD'^FK'. 

Proof.  In  truncated  prisms  AK  and  A'K',  faces  FGHKL 
and  F'G'H'K'L'  are  equal.  (§  475) 

Therefore,  A'K'  may  be  applied  to  AK  so  that  vertices 
F',  G',  etc.,  shall  fall  at  F,  G,  etc.,  respectively. 

Then,  edges  A'F',  B'G',  etc.,  will  coincide  in  direction 
with  AF,  BG,  etc.,  respectively.  (§  399) 

But  since,  by  hyp.,  FF'  =  AA',  we  have  AF  =  A'F\ 

In  like  manner,  BG  =  B'G',  CH=  C'H',  etc. 

Hence,  vertices  A',  B',  etc.,  will  fall  at  A,  B,  etc.,  respec- 
tively. 

Then,  A'K'  and  ^j^  coincide  throughout,  and  are  equal. 

Now  taking  from  the  entire  solid  AK'  truncated  prism 
A'K',  there  remains  prism  AD'. 

And  taking  its  equal  AK,  there  remains  prism  FK'. 
.'.  AD'^FK. 


278 


SOLID   GEOMETRY.— BOOK  VII. 


Prop.  IV.     Theorem. 

481.    The  opposite   lateral  faces  of  a  parallelopiped   are 
equal  and  parallel. 

D'  c' 


Given  AC  and  A'C  the  bases  of  parallelopiped  AC. 
To  Prove     faces  AB'  and  DC  equal  and  11. 
Proof.  AB  is  equal  and  II  to  DC,  and  AA'  to  DD'.  (§  106, 1) 
.-.  Z  A'AB  =  Z  D'DC,  and  AB'  11  DC.       (§  426) 
.-.  face  AB'  =  face  DC.  (§  113) 

Similarly,  we  may  prove  AD'  and  BC  equal  and  II. 

482.  Cor.  Either  face  of  a  parallelopiped  may  be  taken  as 
the  base. 

Prop.  V.     Theorem. 

483.  The  plane  passed  through  tico  diagonally  opposite 
edges  of  a  parallelopiped  divides  it  into  two  equivalent  tria^i- 
gular  prisms. 


Given  plane  AC  passing  through  edges  A  A'  and  CC  of 
parallelopiped  A'C. 

To  Prove     prism  ABC-A'  =o  prism  ACD-A'. 


PRISMS   AND  PARALLELOPIPEDS.  279 

Proof.  Let  EFGH  be  a  right  section  of  the  parallelopiped, 
intersecting  plane  AA'C'Cin  line  EG. 

Now,  face  AB'  II  face  DC.  (§  481) 

.-.  EFW  GH  .  (§  414) 

In  like  manner,  EH  II  FG,  and  EFGH  is  a  O. 

.-.  A  EFG  =  A  ^(^jy.  (§  108) 

Now,  ABC-A  is  =0=  a  right  prism  whose   base  is  EFG 

and  altitude  AA\  and  ACD-A'  is  =0=  a  right  prism  whose 

base  is  EGH  and  altitude  AA\  (§  480) 

But  these  right  prisms  are  equal,  for  they  have  equal 

bases  and  the  same  altitude.  (§  478) 

.-.  ABC-A^  ^  ACD-A\ 

Prop.  VI.     Theorem. 

484.    Tlie  lateral  area  of  a  prism  is  equal  to  the  perimeter 
of  a  right  section  multiplied  by  a  lateral  edge. 


Given  DEFGH  a  right  section  of  prism  AO. 

To  Prove    lat.  area  AC  =  (DE  +  EF -\-  etc.)  x  AA'. 

Proof.     We  have,  AA'  ±  DE.  (§  398) 

.-.  area  AA'B'B  =  DE  x  AA'.  (§  309) 

Similarly,         area  BB'CO  =  EF  x  BB' 

=  EFxAA';  etc.         (§468) 

Adding  these  equations,  we  have 

lat.  area  AC  =  DE  x  AA'  +  EF  x  AA'  +  etc. 
=  {DE  -\-EF+  etc.)  x  AA'. 


280 


SOLID   GEOMETRY.  — BOOK   VII. 


485.  Cor.     The  lateral  area  of  a  right  prism  is  equal  to 
the  perimeter  of  the  base  m^dtiplied  by  the  altitude. 

Prop.  VII.     Theorem. 

486.  Two  rectangidar  parallelopipeds  having  equal  bases 
are  to  each  other  as  their  altitudes. 

Note.     The  phrase  "rectangular  parallelepiped "  in  the  above 
statement  signifies  the  volume  of  the  rectangular  parallelopiped. 

Case  I.     When  the  altitudes  are  commensurable. 


^V^ 


c^ 


/l--^ 

/    yT 

/     / 

/     / 

-'i 

/  / 

/ 

/ 

Given  P  and  Q  rect.  parallelopipeds,  with  equal  bases, 
and  commensurable  altitudes,  AAl  and  BB\ 

Proof.     Let  AC  be  a  common  measure  of  AA^  and  BB\ 

and  suppose  it  to  be  contained  4  times  in  AA!,  and  3  times 

in  BB\ 

.  ^  =  1 
* '  BB'     3* 


(1) 


Through  the  several  points  of  division  of  AA  and  BB' 
pass  planes  ±  to  lines  AA'  and  BB\  respectively. 

Then,  rect.  parallelopiped  P  will  be  divided  into  4  parts, 
and  rect.  parallelopiped  Q  into  3  parts,  all  of  which  parts 

(§  478) 

(2) 


(?) 


will  be  equal. 

P     4 

From  (1)  and  (2), 

P     AA' 
Q      BB' 

PEISMS   AND   PARALLELOPIPEDS. 


281 


Case  II.      Wlien  the  altitudes  are  incommensurable. 


a' 

P 

A 

/ 

L 

A 

/ 

/ 

Given  P  and  Q  rect.  parallelepipeds,  with  equal  bases,  and 
incommensurable  altitudes,  AA  and  BB\ 

To  Prove  ^  =  M. 

Q     BW 

Proof.  Divide  AA}  into  any  number  of  equal  parts,  and 
apply  one  of  these  parts  to  BB^  as  a  unit  of  measure. 

Since  AA^  and  BB^  are  incommensurable,  a  certain  num- 
ber of  the  parts  will  extend  from  B  to  (7,  leaving  a  remainder 
CB'  <  one  of  the  parts. 

Draw  plane  CDJLBB',  and  let  rect.  parallelepiped  BD 
be  denoted  by  Q'. 

Then  since,  by  const.,  AA'  and  BC  are  commensurable, 

1  =  ^-  (§486,  Case  I) 

Kow  let  the  number  of  subdivisions  of  AA'  be  indefinitely 
increased. 

Then  the  length  of  each  part  will  be  indefinitely  dimin- 
ished, and  remainder  CB'  will  approach  the  limit  0. 

P  P 

—  will  approach  the  limit  — > 


Then, 


and 


AA 


AA' 


^^==—  will  approach  the  limit  — — 


BC 


BB' 


AA' 
BB'' 


(§  188) 


487.   Def.     The   dimensions  of  a  rectangular  parallelo- 
piped  are  the  three  edges  which  meet  at  any  vertex. 


282 


SOLID   GEOMETRY.  — BOOK  VII. 


488.   Sch.     The  theorem  of  §  486  may  be  expressed : 

If  two  rectangular  parallelopipeds  have  two  dimensions  of 

one  equal  respectively  to  tivo  dimensions  of  the  other,  they  are 

to  each  other  as  their  third  dimensions. 


Prop.  VIII.     Theorem. 

489.    Two  rectangular  parallelopipeds  having  equal  altitudes 
are  to  each  other  as  their  bases. 


p 

Q 

/ 

/ 

A 

/ 

/ 

/ 

A 

/ 

c 

/ 

c 

^j — 

p_ 

/ 

/ 

\x 

Given  P  and  Q  rect.  parallelopipeds,  with  the  same  alti- 
tude c,  and  the  dimensions  of  the  bases  a,  6,  and  a',  h\ 
respectively. 

P      axb 


To  Prove 


(§  305) 


Q      a'  xb' 

Proof.     Let  i?  be  a  rect.  parallelopiped  with  the  altitude 
c,  and  the  dimensions  of  the  base  a'  and  b. 

Then  since  P  and  M  have  each  the  dimensions  b  and  c, 
they  are  to  each  other  as  their  third  dimensions  a  and  a'. 

(§  488) 

That  is,  ?:  =  -.•  (1) 


E     a' 
And  since  R  and  Q  have  each  the  dimensions  a'  and  c, 

R^b 

Q     b' 

Multiplying  (1)  and  (2),  we  have 


(2) 


R 


or 


axb 


R      Q'        Q     a'  xb' 


PRISMS  AND  PARALLELOPIPEDS. 


283 


490.   Sch.     The  theorem  of  §  489  may  be  expressed : 
Two  rectangular  parallelopipeds  having  a  dimeyision  of  one 

equal  to  a  dimension  of  the  other,  are  to  each  other  as  the 

products  of  their  other  two  dimensions. 


Prop.  IX.     Theorem. 

491.  Any    two    rectangular  parallelopipeds   are   to   each 
other  as  the  products  of  their  three  dimensions. 


p  R 

/~7\  Ay\  47 

i  f- — W  I 

/ /  /      '  /  /       / 

A  /      /^  /       A 

k 1/  k 1/  k: 1/ 


Given  P  aud  Q  rect.  parallelopipeds  with  the  dimensions 
a,  b,  c,  and  a',  6',  c',  respectively. 

To  Prove  ^^  ?  \^/ V 

Q     a'  X  b'  X  c' 

(Let  I{  be  a  rect.  parallelopiped  with  the  dimensions  a', 
b',  and  c,  and  find  values  of  ^  and  :?  by  §§  490  and  488.) 


EXERCISES. 

1.  Two  rectangular  parallelopipeds,  with  equal  altitudes,  have  the 
dimensions  of  their  bases  6  and  14,  and  7  and  9,  respectively.  Find 
the  ratio  of  their  volumes. 

2.  Find  the  ratio  of  the  volumes  of  two  rectangular  parallelopipeds, 
whose  dimensions  are  8,  12,  and  21,  and  14,  15,  and  24,  respectively. 

D_'     _C' 

3.  The  diagonals  of  a  parallelopiped  bisect  each 
other. 

(To  prove  that  AO  and  ^'C  bisect  each  other. 
Prove  AA'C'Gz.  O  by  §  110.) 


284 


SOLID   GEOMETRY.— BOOK  VII. 


Prop.  X.     Theorem. 

492.  If  the  unit  of  volume  is  the  cube  whose  edge  is  the 
linear  unit,  the  volume  of  a  rectangular  parallelopiped  is 
equal  to  the  product  of  its  three  dimensions. 


/ X 


A 

/ 

\ 
/ 

1 

A 

Given  a,  6,  and  c  the  dimensions  of  rect.  parallelopiped 
P,  and  Q  the  unit  of  volume ;  that  is,  a  cube  whose  edge 
is  the  linear  unit. 

To  Prove  vol.  P=a  xh  xc. 

P _  g-x  b  X  c 


Proof.     We  have 


1x1x1 
=  a  X  6  X  c. 
But  since  Q  is  the  unit  of  volume, 

"=  vol.  P. 


(§  491) 


Q 

vol.  P=  a  X  b  X  c. 


(§  464) 


493.  Sch.  I.  In  all  succeeding  theorems  relating  to  vol- 
umes, it  is  understood  that  the  unit  of  volume  is  the  cube 
whose  edge  is  the  linear  unit,  and  the  unit  of  surface  the 
square  whose  side  is  the  linear  unit.     (Compare  §  306.) 


494.   Cor.  I. 
of  its  edge. 


The  volume  of  a  cube  is  equal  to  the  cube 


495.   Cor.  11!     TJie  volume  of  a  rectangular  parallelopiped 
is  equal  to  the  product  of  its  base  and  altitude. 
(The  proof  is  left  to  the  pupil.) 


PRISMS   AND  PARALLELOPIPEDS. 


285 


496.   Sch.  II.     If    the    dimensions    of    the    rectangular 
parallelopiped  are  tmdtiples  of  the  linear 
unit,  the  truth  of  Prop.  X.  may  be  seen  P 

by   dividing  the   solid  into  cubes,  each 
equal  to  the  unit  of  volume. 

Thus,  if  the  dimensions  of  rectangular 
parallelopiped  P  are  5  units,  4  units,  and 
3  units,  respectively,  the  solid  can  evi- 
dently be  divided  into  60  cubes. 

In  this  case,  60,  the  number  which  expresses  the  volume 
of  the  rectangular  parallelopiped,  is  the  product  of  5,  4,  and 
3,  the  numbers  which  express  the  lengths  of  its  edges. 


r — r — t — i' — 

._.!._!     ! 

1     1     1 

yi. 


EXERCISES. 

4.  Find  the  altitude  of  a  rectangular  parallelopiped,  the  dimen- 
sions of  whose  base  are  21  and  30,  equivalent  to  a  rectangular  paral- 
lelopiped whose  dimensions  are  27,  28,  and  35. 

5.  Find  the  edge  of  a  cube  equivalent  to  a  rectangular  parallelo- 
piped whose  dimensions  are  9  in.,  1  ft.  9  in.,  and  4  ft.  1  in. 

6.  Find  the  volume,  and  the  area  of  the  entire  surface  of  a  cube 
whose  edge  is  3^  in. 

7.  Find  the  area  of  the  entire  surface  of  a  rectangular  parallelo- 
piped, the  dimensions  of  whose  base  are  11  and  13,  and  volume  858. 

8.  Find  the  volume  of  a  rectangular  parallelopiped,  the  dimen- 
sions of  whose  base  are  14  and  9,  and  the  area  of  whose  entire  surface 
is  620. 

9.  Find  the  dimensions  of  the  base  of  a  rectangular  parallelo- 
piped, the  area  of  whose  entire  surface  is  320,  volume  336,  and 
altitude  4. 

(Represent  the  dimensions  of  the  base  by  x  and  y. ) 

10.   How  many  bricks,  each  8  in.  long,  2f  in.  wide,  and  2  in. 

thick,  will  be  required  to  build  a  wall  18  ft.  long,  3  ft.  high,  and 

11  in.  thick  ? 


11.   The  diagonals  of  a  rectangular  parallelo- 
piped are  equal. 

(Prove  A  A*  C"  C  a  rectangle. ) 


286 


SOLID   GEOMETRY.  — BOOK  VII. 


Prop.  XI.     Theorem. 

497.    Tlie   volume  of  any  parallelopiped   is   equal   to  the 
product  of  its  base  and  altitude. 


Given  AE  the  altitude  of  parallelopiped  AC. 

To  Prove  vol.  AC  =  area  ABCD  x  AE. 

Proof.     Produce  edges  AB,  A'B\  D'C,  and  DO. 

On  AB  produced,  take  FG  =  AB ;  and  draw  planes  FK' 
and  GH'  J_  FG,  forming  right  parallelopiped  FH'. 

.-.  FH'^AC.  (§480) 

Produce  edges  HG,  H'G',  K'F,  and  KF. 
On  HG  produced,  take  NM=  HG;  and  draw  planes  iVP' 
and  ML'  ±  NM,  forming  right  parallelopiped  LN'. 

.'.  LN'^FH'.  (§  480) 

.-.  LN'^AC. 

Now  since,  by  cons.,  FG  is  _L  plane  GH',  planes  LH  and 

MH'  are  ±.  (§  441) 

Then  MM',  being  J.  MN,  is  ±  plane  LH.  (§  438) 

Whence,  Z  LMM'  is  a  rt.  Z.  (§  398) 

Then,  LM'  is  a  rectangle.  (§  76) 
Therefore  iiV'  is  a  rectangular  parallelopiped. 

.-.  vol.  LN'  =  area  LMNP  x  JOf' .  (§  495) 

.-.  vol.  AC  =  area  LMNP  x  JO/'.  (1) 


PRISMS  AND  PARALLELOPIPEDS.  287 

But  rect.  LMNP  =  rect.  FGHK\   for  they  have  equal 

bases  MN  and  GH,  and  the  same  altitude.  (§  114) 

Also,  rect.  FOHK^nABCD  ;  for  they  have  equal  bases 

FG  and  AB^  and  the  same  altitude.  (§  310) 

.-.  LMNP^ABCD. 

Also,  JOf '  =  AE.  (§  424) 

Substituting  these  values  in  (1),  we  have 

vol.  AC  =  area  ABCD  x  AE.      ' 

Prop.  XII.     Theorem. 

498.    The   volume   of  a  triangular  prism   is  equal  to  the 
product  of  its  base  and  altitude. 


Given  AE  the  altitude  of  triangular  prism  ABC-O. 
To  Prove      vol.  ABC-O  =  area  ABC  x  AE. 
Proof.     Construct   parallelopiped   ABCD-D',  having  its 
edges  II  to  AB,  BC,  and  BB',  respectively. 

.-.  vol.  ABC-C  =  1  vol.  ABCD-D'  (§  483) 

=  i-  area  ABCD  x  AE      (§  497) 
=  area  ABC  X  AE.  (§  108) 

EXERCISES. 

12.  Find  the  lateral  area  and  volume  of  a  regular  triangular  prism, 
each  side  of  whose  base  is  5,  and  whose  altitude  is  8. 

13.  The  square  of  a  diagonal  of  a  rectangular  parallelopiped  is 
equal  to  the  sum  of  the  squares  of  its  dimensions. 

(Fig.  of  Ex.  11.     To  prove  JJC^  =  AA^  +  AB^  +  AD\) 


288  SOLID   GEOMETRY. —BOOK  VII. 


Pkop.  XIII.     Theorem. 

499.    The  volume  of  any  prism  is  equal  to  the  product  of  its 
base  and  altitude. 


Given  any  prism. 

To  Prove  its  volume  equal  to  the  product  of  its  base  and 
altitude. 

Proof.  The  prism  may  be  divided  into  triangular  prisms 
by  passing  planes  through  one  of  the  lateral  edges  and  the 
corresponding  diagonals  of  the  base. 

The  volume  of  each  triangular  prism  is  equal  to  the  prod- 
uct of  its  base  and  altitude.  (§  498) 

Then,  the  sum  of  the  volumes  of  the  triangular  prisms  is 
equal  to  the  sum  of  their  bases  multiplied  by  their  common 
altitude. , 

Therefore,  the  volume  of  the  given  prism  is  equal  to  the 
product  of  its  base  and  altitude. 

500.  Cor.  I.  Two  prisms  having  equivalent  bases  and 
equal  altitudes  are  equivalent. 

501.  Cor.  IT.  1.  Two  prisms  having  equal  altitudes  are 
to  each  other  as  their  bases. 

2.  Two  prisms  having  equivalent  bases  are  to  each  other  as 
their  altitudes. 

3.  Any  two  prisms  are  to  each  other  as  the  products  of  their 
bases  by  their  altitudes. 


Ex.  14.   Find  the  lateral  area  and  volume  of  a  regular  hexagonal 
prism,  each  side  of  whose  base  is  3,  and  whose  altitude  is  9. 


PYRAMIDS. 


289 


PYRAMIDS. 
DEFINITIONS. 

502.  A  pyramid  is  a  polyedron  bounded  by  a  polygon, 
called  the  base,  and  a  series  of  triangles 

having  a  common  vertex. 

The  common  vertex  of  the  triangular 
faces  is  called  the  vertex  of  the  pyramid. 

The  triangular  faces  are  called  the  lateral 
faces,  and  the  edges  terminating  at  the  vertex 
the  lateral  edges. 

The  sum  of  the  areas  of  the  lateral  faces  is  called  the 
lateral  area. 

The  altitude  is  the  perpendicular  distance  from  the  vertex 
to  the  plane  of  the  base. 

503.  A  pyramid  is  called  triangular,  quadrangular,  etc., 
according  as  its  base  is  a  triangle,  quadrilateral,  etc. 

504.  A  regular  pyramid  is  a  pyramid  whose 
base  is  a  regular  polygon,  and  whose  vertex 
lies  in  the  perpendicular  erected  at  the  centre 
of  the  base. 

505.  A  truncated  pyramid  is  a  portion  of 
a  pyramid  included  between  the  base  and  a  plane  cutting 
all  the  lateral  edges. 

The  base  of  the  pyramid  and  the  section  made  by  the 
plane  are  called  the  bases  of  the  truncated  pyramid. 

506.  A  frustum  of  a  pyramid  is  a  trun- 
cated pyramid  whose  bases  are  parallel. 

The  altitude  is  the  perpendicular  distance 
between  the  planes  of  the  bases. 

EXERCISES. 

15.  Find  the  length  of  the  diagonal  of  a  rectangular  parallelepiped 
whose  dimensions  are  8,  9,  and  12, 

16.  The  diagonal  of  a  cube  is  equal  to  its  edge  multiplied  by  V3. 


290 


SOLID   GEOMETRY.— BOOK  VII. 


Prop.  XIV.     Theorem. 
507.   In  a  regular  pyramid, 
The  lateral  edges  are  equal. 
The  lateral  faces  are  equal  isosceles  triangles. 
o 


I. 
11. 


B  C 

(The  theorem  follows  by  §§  406,  I,  and  69.) 

508.  Def.     The  slant  height  of  a  regular  pyramid  is  the 
altitude  of  any  lateral  face. 

Or,  it  is  the  line  drawn  from  the  vertex  of  the  pyramid  to 
the  middle  point  of  any  side  of  the  base.  (§  94,  I) 

Prop.  XV.     Theorem. 

509.  Tlie  lateral  faces  of  a  frustum  of  a  regular  pyramid 
are  equal  trapezoids.  ^ 


B  C 

Given  AC  a  frustum  of  regular  pyramid  0-ABCDE. 

To  Prove  faces  AB'  and  EC  equal  trapezoids. 

Proof.     We  have    A  OAB  =  A  OBC.  (§  507,  II) 

We  may  then  apply  A  OAB  to  A  OBC  in  such  a  way 
that  sides  OB,  OA,  and  AB  shall  coincide  with  sides  OB, 
00,  and  BC,  respectively. 


PYRAMIDS.  291 

Now,  A'B'  II  AB  and  B'C  II  BC.  (?) 

Hence,  line  A'B'  will  coincide  with  line  B'C.  (§  53) 

Then,  AB'  and  BC  coincide  throughout,  and  are  equal. 

510.  Cor.     The  lateral  edges  of  a  frustum  of  a  regular 
pyramid  are  equal. 

511.  Def.     The  slant  height  of  a  frustum  of  a  regular 
pyramid  is  the  altitude  of  any  lateral  face. 

Prop.  XVI.     Theorem. 

512.  The  lateral  area  of  a  regular  pyramid  is  equal  to  the 
perimeter  of  its  base  multiplied  by  one-half  its  slant  height. 

O 


B  C 

Given  slant  height  OH  of  regular  pyramid  0-ABCDE. 
To  Prove 

lat.  area  0-ABCDE  =  (AB  -\- BC -{-  etc.)  x  i  OH. 
(By  §  508,  OH  is  the  altitude  of  each  lateral  face.) 

513.   Cor.    The   lateral  area  of  a  frustum   of  a  regidar 
pyramid  is  equal  to  one-half  the  sum  of 
the  perimeters  of  its  bases,  multiplied  by         A'^-'-rr—-~-yJ)' 
its  slant  height.  / -Ib^I — r\ 

Given  slant  height  HH'  of  the  frus-    ^/-/^/'^"   y"--4jD 
tum  of  a  regular  pyramid  AD'.  h\L \X 

B  C 

To  Prove 

lat.  area  AD'  =  \  {AB  +  A'B'  +  BC+B'C  -f  etc.)  x  HH', 
{HH  is  the  altitude  of  each  lateral  face.) 


292  SOLID   GEOMETRY.  — BOOK  VIT. 


EXERCISES. 

17.  The  volume  of  a  cube  is  4^^  cu.  ft.  Find  the  area  of  its  entire 
surface  in  square  inches. 

18.  The  volume  of  a  right  prism  is  2310,  and  its  base  is  a  right 
triangle  whose  legs  are  20  and  21,  respectively.     Find  its  lateral  area, 

19.  Find  the  lateral  area  and  volume  of  a  right  triangular  prism, 
having  the  sides  of  its  base  4,  7,  and  9,  respectively,  and  the  altitude  8. 

20.  The  volume  of  a  regular  triangular  prism  is  96  Vs,  and  one 
side  of  its  base  is  8.     Find  its  lateral  area. 

21.  The  diagonal  of  a  cube  is  8  VS.  Find  its  volume,  and  the 
area  of  its  entire  surface. 

(Represent  the  edge  by  x.) 

22.  A  trench  is  124  ft.  long,  2|  ft.  deep,  6  ft.  wide  at  the  top,  and 
5  ft.  wide  at  the  bottom.  How  many  cubic  feet  of  water  will  it  con- 
tain ?     (§§  316,  499.) 

23.  The  lateral  area  and  volume  of  a  regular  hexagonal  prism  are 
60  and  15  V3,  respectively.     Find  its  altitude,  and  one  side  of  its  base. 

(Represent  the  altitude  by  x,  and  the  side  of  the  base  by  y.) 

Prop.  XVII.     Theorem. 

514.   If  a  pyramid  he  cut  by  a  plane  parallel  to  its  6ase, 
1.    The  lateral  edges  and  the  altitude  are  divided  propor- 
tionally. 

II.    The  section  is  similar  to  the  base. 


Given  plane  ^'C  II  to  base  of  pyramid  0-ABCD,  cutting 
faces  OAB,  OBC,  OCD,  and  ODA  in  lines  A^B\  B'C,  CD', 
and  D'A',  respectively,  and  altitude  OP  at  P'. 


PYRAMIDS. 


I.   To  Prove    ^^  =  M' =  ^' etc.  =  ^. 
OA       OB       OC  OP 

Proof.    Through  0  pass  plane  MN  II  ABCD. 

OA^OB^^qC         _0P' 
"  OA      OB      00  op' 


293 


(§  427) 


II.   To  Prove  section  A'B'C'D'  similar  to  ABCD. 

Proof.     We  have  A'B'  II  AB,  B'C  II  BC,  etc.  (?) 

.-.  Z  A'B'C  =  Z  ABC,  Z  BOU  =  ZBCD,  etc.  (§  426) 

Again,  A  OA^B\  OB'C,  etc.,  are  similar  to  A  OAB,  OBC, 

etc.,  respectively.  (§  257) 

OA'     A'B'     OB'     B'C 


OA       AB'    OB       BC 


etc.  (1) 


AB       BC       CD  ^  '^ 

Then,  polygons  A'B'C'D'  and  ABCD  are  mutually  equi- 
angular, and  have  their  homologous  sides  proportional. 
Whence,  A'B'C'D'  and  ABCD  are  similar.  (§  252) 


515.   Cor.  I.     Since  A'B'C'D'  and  ABCD  are  similar, 
area  A'B'C'D'     A^'' 


But  from  (1),  §  514, 


area  ABCD 

AB' 

rii4. 

A'B' 

OA' 

AB 

OA 
OP' 
OP 

area 

A'B'C'D' 

OP' 

(§  322) 


(§514,1) 


area  ABCD        Qp^ 

Hence,  the  area  of  a  section  of  a  pyramid,  parallel  to  the 
base,  is  to  the  area  of  the  base  as  the  square  of  its  distance 
from  the  vertex  is  to  the  square  of  the  altitude  of  the  pyramid. 


294 


SOLID   GEOMETRY.— BOOK  VII, 


516.   Cor.  II.     If  two  pyramids  have  equal  altitudes  and 
equivalent    bases,    sections 
parallel  to  the  bases  equally 
distant  from    the  vertices 
are  equivalent. 

Given  bases  of  pyramids 
0-^50  and  0'-^'^'(7'=o, 
and  the  altitude  of  each 
pyramid  =  fi";  also  DEF 
and  D'E'F'  sections  =  to 
the  bases  at  distance  h  from  0  and  0',  respectively. 

To  Prove  area  DEF  =  area  D'E'F'. 

Proof.     We  have 

area  DEF^  h^    ^^^a^vesi  D'E'F'  ^  W    /.k^kx 
area^SC     H^'         aTesiA'B'C      ^^'  ^         ^ 


W 


But  by  hyp., 


area  DEF  _  area  D'E'F' 
area  ABC  ~  area  A'B'O  ' 
area  ABC  =  area  A'B'C. 
area  DEF=  area  D'E'F'. 


(?) 


Prop.  XVIII.     Theorem. 

517.    Two  triangular  pyramids  having  equal  altitudes  and 
equivalent  bases  are  equivalent. 


Given  o-abc  and  o'-a'b'c'  triangular  pyramids  with  equal 
altitudes  and  =c=  bases. 


PYRAMIDS. 


295 


To  Prove  vol.  o^abc  =  vol.  o'-a%'c\ 

Proof.  Place  the  pyramids  with  their  bases  in  the  same 
plane,  and  let  PQ  be  their  common  altitude. 

Divide  PQ  into  any  number  of  equal  parts. 

Through  the  points  of  division  pass  planes  II  to  the  plane 
of  the  bases,  cutting  o-abc  in  sections  def  and  ghk,  and 
O'-a'b'c'  in  sections  d'e'f  and  g'h'k',  respectively. 

.-.  def^  d'e'f,  and  ghk  ^  g'h'k'.  (§  516) 

With  abc,  def,  and  ghk  as  lowei'  bases,  construct  prisms 
X,  Y,  and  Z,  with  their  lateral  edges  equal  and  II  to  ad; 
and  with  d'e'f  and  g'h'k'  as  upper  bases,  construct  prisms 
T'  and  Z',  with  their  lateral  edges  equal  and  II  to  a'd'. 

.*.  prism  Y^  prism  Y',  and  prism  Zo  prism  Z'.  (§  500) 

Hence,  the  sum  of  the  prisms  circumscribed  about  o-abc 
exceeds  the  sum  of  the  prisms  inscribed  in  o'-a'b'c'  by 
prism  X 

But,  o-abc  is  evidently  <  the  sum  of  prisms  X,  Y,  and 
Z;  and  it  is  >  the  sum  of  prisms  o=  to  Y'  and  Z',  respec- 
tively, which  can  be  constructed  with  def  and  ghk  as  upper 
bases,  having  their  lateral  edges  equal  and  II  to  ad. 

Again,  o'-a'b'c'  is  >  the  sum  of  prisms  Y'  and  Z' ;  and 
it  is  <  the  sum  of  prisms  =c=  to  X,  Y,  and  Z,  respectively, 
which  can  be  constructed  with  a'b'c',  d'e'f,  and  g'h'k'  as 
lower  bases,  having  their  lateral  edges  equal  and  II  to  a'd'. 

That  is,  each  pyramid  is  <  the  sum  of  prisms  X,  Y,  and 
Z,  and  >  the  sum  of  prisms  Y'  and  Z' ;  whence,  the  differ- 
ence of  the  volumes  of  the  pyramids  must  be  <  the  dif- 
ference of  the  volumes  of  the  two  systems  of  prisms, 
or  <  volume  X. 

Now  by  sufficiently  increasing  the  number  of  subdivisions 
of  PQ,  the  volume  of  prism  X  may  be  made  <  any  assigned 
volume,  however  small. 

Hence,  the  volumes  of  the  pyramids  cannot  differ  by  any 
volume,  however  small. 

.•.  vol.  o-abc  =  vol.  o'-a'b'c'. 


296  SOLID   GEOMETRY.— BOOK   VII. 

518.  Cor.  Since  vol.  o'-a'b'c'  is  >  the  total  volume  of 
the  inscribed  prisms,  and  <  the  total  volume  of  the  cir- 
cumscribed, the  difference  between  vol.  o'-a'b'c'  and  the 
total  volume  of  the  inscribed  prisms  is  <  the  difference 
between  the  total  volumes  of  the  two  systems  of  prisms, 
or  <  vol.  X]  and  hence  approaches  the  limit  0  when  the 
number  of  subdivisions  is  indefinitely  increased. 

Prop.  XIX.     Theorem. 

519.  A  triangular  pyramid  is  equivalent  to  one-third  of  a 
triangular  prism  having  the  same  base  and  altitude. 


Given  triangular  pyramid  0-ABC,  and  triangular  prism 
ABC-ODE  having  the  same  base  and  altitude. 

To  Prove      vol.  0-ABC  =  i  vol.  ABC-ODE. 

Proof.  Prism  ABC-ODE  is  composed  of  triangular  pyra- 
mid 0-ABC,  and  quadrangular  pyramid  0-ACDE. 

Divide  the  latter  into  two  triangular  pyramids,  0-ACE 
and  0-CDE,  by  passing  a  plane  through  0,  C,  and  E. 

Now,  0-ACE  and  0-CDE  have  the  same  altitude. 

And  since  CE  is  a  diagonal  of  O  ACDE,  they  have  equal 
bases,  ACE  and  CDE.  (§  108) 

.-.  vol.  0-ACE  =  vol.  0-CDE.  (§  517) 

Again,  pyramid  0-CDE  may  be  regarded  as  having  its 
vertex  at  C,  and  A  ODE  for  its  base. 

Then,  pyramids  0-ABC  and  C-ODE  have  the  same 
altitude.  (§  424) 


PYRAMIDS.  297 

They  have  also  equal  bases,  ABC  and  ODE.  (§  467) 

.-.  vol.  0-ABC=  vol  C-ODE.  (?) 

Then,  vol.  0-ABG  =  vol.  0-ACE  =  vol.  0-CDE.         (?) 
.-.  vol.  0-ABC  =  ^  vol.  ABC-ODE. 

520.  Cor.     The  volume  of  a  triangular  pyramid  is  equal  to 
one-third  the  product  of  its  base  and  altitude.  (§  498) 

Prop.  XX.     Theorem. 

521.  The  volume  of  any  pyramid  is  equal  to  one-third  the 
product  of  its  base  and  altitude. 


(Prove  as  in  §  499.) 

522.  Cor.  1.  Two  pyramids  having  equivalent  bases  and 
equal  altitudes  are  equivalent. 

2.  Two  pyramids  having  equal  altitudes  are  to  each  other 
as  their  bases. 

3.  Two  pyramids  having  equivalent  bases  are  to  each  other 
as  their  altitudes. 

4.  Any  two  pyramids  are  to  each  other  as  the  products  of 
their  bases  by  their  altitudes. 

EXERCISES. 

24.  The  altitude  of  a  pyramid  is  12  in.,  and  its  base  is  a  square 
9  in.  on  a  side.  What  is  the  area  of  a  section  parallel  to  the  base, 
whose  distance  from  the  vertex  is  8  in.  ?     (§  515.) 

25.  The  altitude  of  a  pyramid  is  20  in.,  and  its  base  is  a  rectangle 
whose  dimensions  are  10  in.  and  15  in.,  respectively.  What  is  the  dis- 
tance from  the  vertex  of  a  section  parallel  to  the  base,  whose  area  is 
54  sq.  in. ? 


298 


SOLID   GEOMETRY.— BOOK  VII. 


Prop.  XXI.     Theorem. 

523.  Two  tetraedro7is  hav'my  a  triedral  angle  of  one  equal 
to  a  triedral  angle  of  the  other,  are  to  ea^h  other  as  the  products 
of  the  edges  including  the  equal  triedral  angles. 

C 


Given  V  and  F'  the  volumes  of  tetraedrons  0-ABO  and 
O-A'B'C,  respectively,  having  the  common  triedral  Z  0. 

To  Prove  Z=  OA  x  OB  x  PC  ^ 

V     OA'  X  OB'  X  OC 

Proof.     Draw  lines  CP  and  C'P'  ±  to  face  OA'B'. 
Let  their  plane  intersect  face  OA'B'  in  line  OPP'. 
■  Now,  OAB  and  OA'B'  are  the  bases,  and  CP  and  CP 
the  altitudes,  of  triangular  pyramids  C-OAB  and  C'-OA'B', 
respectively. 

V  ^   area  OAB  x  CP 
"  V 


area  OA'B'  x  CP' 
area  OAB       CP 


X 


area  OA'B'     CP 
But  area  OAB  _  OA  x  OB 

'  3ivesi,0A'B'~  OA'  x  OB'' 

Also,  A  OCP  and  OCP'  are  rt.  A. 
Then,  A  OCP  and  OCP  are  similar. 
CP  ^OC 
"  CP'     oc' 
Substituting  these  values  in  (1),  we  have 


(§  ^^^2,  4) 

(1) 
(§  321) 


398) 
256) 

(?) 


_F 

V 


OA  X  OB  ..  OC       OA  X  OBx  OC 


OA'  X  OB'     OC      OA'  X  OB'  x  OC 


PYRAMIDS.  299 


Prop.  XXII.     Theorem. 

524.  The  volume  of  a  frustum  of  a  pyramid  is  equal  to  the 
sum  of  its  bases  and  a  mean  proportional  between  its  bases, 
multiplied  by  one-third  its  altitude. 

O 


B 

Given  B  the  area  of  the  lower  base,  b  the  area  of  the 
upper  base,  and  ^  the  altitude,  of  AC,  a  frustum  of  any 
pyramid  0-AC. 
To  Prove  vol.  AC  =(B  +  b+  VJB  x  b)  x^H.      (§  233) 
Proof.     Draw  altitude  OP,  cutting  A'C  at  Q. 
Now,    vol.  AC  =  vol.  0-AG  -  vol.  0-A'C 

=  Bx  K^+  OQ)-bx  iOQ       (§  521) 
=  BxlH+BxiOQ-bxiOQ 
=  Bx^H+{B-b)x\Oq.  (1) 

But,  B:b  =  OF':OQ\  (§515) 

Taking  the  square  root  of  each  term, 

^B:^b=OP'.OQ.  (§241) 

...  VB-Vb:Vb=OP-OQ:OQ         (§238) 
=  H:OQ. 
.'.  (VB  -  Vb)  xOQ=VbxH.  (§  232) 

Multiplying  both  members  by  ( V^  +  V^), 

(B-b)  xOQ=  (V^x  b  -\-b)  xH. 
Substituting  this  value  in  (1),  we  have 

vol.  AC  =  BxiII+  ( VS^Tft  -\-b)xiH 
=  (B  +  b+VWxb)xiH, 


300  SOLID   GEOMETRY.— BOOK  VII. 

Prop.  XXIII.     Theorem. 

525.  The  volume  of  a  truncated  tria7igular  prism  is  equal 
to  the  product  of  a  right  section  by  one-third  the  sum  of  the 
lateral  edges.  _P 


B 

Given  GHC  and  DKL  rt.  sections  of  truncated  triangular 
prism  ABC-DEF. 
To  Prove 

vol.  ABC-DEF  =  area  GHC  x  \  {AD  +  BE  +  CF). 
Proof.     Draw  line  DM±  KL. 

The  given  truncated  prism  consists  of  the  rt.  triangular 

prism  GHC-DKL,  and  pyramids  D-EKLF  and  C-ABHG. 

vol.  GHC-DKL  =  area  GHC  x  GD  (§  498) 

=  area  GHC  x  \  {GD  +  HK+  CL),  (1) 

since  the  lateral  edges  of  a  prism  are  equal  (§  468). 

Now  DM  is  the  altitude  of  pyramid  D-EKLF.       (§  438) 

.-.  vol.  D-EKLF  =  area  EKLF  x  ^  DM.  (§  521) 

But  KL  is  the  altitude  of  trapezoid  EKLF.  (§  398) 

.-.  vol.  D-EKLF  =  i  {KE  +  LF)  x  KL  x  i  DM.  (§  316) 

Rearranging  the  factors,  we  have 

vol.  D-EKLF  ={\KLx  DM)  x  i  {KE  +  IjF) 

=  area  DKL  x  i  {KE  +  Li^)       (§  312) 
=  area  (7irO  x  ^  (^-E  -f-  i>i^).  (2) 

In  like  manner,  we  may  prove 

vol.  C-ABHG  =  area  GHC  x  i  {AG  +  BH).  (3) 

Adding  (1),  (2),  and  (3),  the  sum  of  the  volumes  of  the 
solids  GHC-DKL,  D-EKLF,  and  C-ABHG  is 
area  GHC  x  ^  {AG  +  GD  +  BH+HK+KE+  CLTLF)- 
.:  vol.  ABC-DEF  =  area  GHC  x  i  {AD  -^  BE  +  CF). 


PYRAMIDS. 


301 


526.  Cor.  Tlie  volume  of  a  truncated  right  triangular 
prism  is  equal  to  the  product  of  its  base  by  one-third  the  sum 
of  the  lateral  edges. 

EXERCISES. 

26.  Each  side  of  the  base  of  a  regular  triangular 
pyramid  is  6,  and  its  altitude  is  4.  Find  its  lateral 
edge,  lateral  area,  and  volume. 

Let  OAB  be  a  lateral  face  of  the  regular  tri- 
angular pyramid,  and  C  the  centre  of  the  base  ; 
draw  line  GDJLAB;  also,  lines  OC,  AC,  and  OD. 


Now,    ^C  =  —  (§  356)  =  A 


2V3. 


lat.  edge  OA  =  ^ AC^  +  0C'\%  272)  =  Vl2  +  16  =  V28  =  2 a/7. 


.-.  slant  ht.  OD  =  Vo^^  -  Alf{%  273)  =  V28  -  9  =  Vl9. 
.-.  lat.  area  of  pyramid  =  9  VlQ  (§  512). 
Again,  CD 


VI^ 


ad'  =  V12-9  =^/Z.  ■ 

.'.  area  of  base  =  ^  x  18  x  V3  (§  350)  =  9\/3. 
.-.  vol.  of  pyramid  =  ^  x  9  VS  x  4  (§  520)  =  12  V3. 

27.  Find  the  lateral  edge,  lateral  area,  and  volume  of  a  frustum  of 
a  regular  quadrangular  pyramid,  the  sides  of  whose  bases  are  17  and 
7,  respectively,  and  whose  altitude  is  12. 

Let  ABB' A'  be  a  lateral  face  of  the  frustum,  and  0  and  0'  the 
centres  of  the  bases;  draw  lines    OC±AB, 
0'C'±A'B',  C'D±  OC,  and  A'E±AB;  also, 
lines  00'  and  CC. 


Now,  CD=OC-  O'C 
.-.  Slant  ht.  CC 


8^-31 


=  V  CD^+  C'D^=  V25  +  144  =  Vl69  =  13. 

.-.  lat,  area  frustum 

=  K68  +  28)  X  13  (§  513)  =  624. 

Again,  AE  =  AC  -  A'C  =  8^  -  3^  =  5,  and  A'E=  CC-  =  13. 

.-.  lat.  edge  AA'  =  Vae^  +  A^'  =  V25  +  169  =  Vl94. 
Again,  area  lower  base  =  17^,  area  upper  base  =  7^,  and  a  mean 
proportional  between  them  =  VlT^  x  T^  =  17  x  7  =  119. 
.-.  vol.  frustum  =(289  +  49  +  119)  x  4  (§  524)  =  1828. 


302  SOLID   GEOMETRY.— BOOK  VII. 

Find  the  lateral  edge,  lateral  area,  and  volume 

28.  Of  a  regular  triangular  pyramid,  each  side  of  whose  base  is  12, 
and  whose  altitude  is  15. 

29.  Of  a  regular  quadrangular  pyramid,  each  side  of  whose  base 
is  3,  and  whose  altitude  is  5. 

30.  Of  a  regular  hexagonal  pyramid,  each  side  of  whose  base  is  4, 
and  whose  altitude  is  9. 

31.  Of  a  frustum  of  a  regular  triangular  pyramid,  the  sides  of 
whose  bases  are  18  and  6,  respectively,  and  whose  altitude  is  24, 

32.  Of  a  frustum  of  a  regular  quadrangular  pyramid,  the  sides  of 
whose  bases  are  9  and  5,  respectively,  and  whose  altitude  is  10. 

33.  Of  a  frustum  of  a  regular  hexagonal  pyramid,  the  sides  of 
whose  bases  are  8  and  4,  respectively,  and  whose  altitude  is  12. 

34.  Find  the  volume  of  a  truncated  right  triangular  prism,  the  sides 
of  whose  base  are  5,  12,  and  13,  and  whose  lateral  edges  are  3,  7,  and 

5,  respectively. 

35.  Find  the  volume  of  a  truncated  right  quadrangular  prism,  each 
side  of  whose  base  is  8,  and  whose  lateral  edges,  taken  in  order,  are  2, 

6,  8,  and  4,  respectively. 

(Pass  a  plane  through  two  diagonally  opposite  lateral  edges,  divid- 
ing the  solid  into  two  truncated  right  triangular  prisms.) 

36.  Find  the  volume  of  a  truncated  right  triangular  prism,  whose 
lateral  edges  are  11,  14,  and  17,  having  for  its  base  an  isosceles  triangle 
whose  sides  are  10,  13,  and  13,  respectively. 

37.  The  slant  height  and  lateral  edge  of  a  regular  quadrangular 
pyramid  are  25  and  V674,  respectively.  Find  its  lateral  area  and 
volume. 

38.  The  altitude  and  slant  height  of  a  regular  hexagonal  pyramid 
are  15  and  17,  respectively.    Find  its  lateral  edge  and  volume. 

(Represent  the  side  of  the  base  by  x.) 

39.  The  lateral  edge  of  a  frustum  of  a  regular  hexagonal  pyramid 
is  10,  and  the  sides  of  its  bases  are  10  and  4,  respectively.  Find  its 
lateral  area  and  volume. 

40.  Find  the  lateral  area  and  volume  of  a  frustum  of  a  regular 
triangular  pyramid,  the  sides  of  whose  bases  are  12  and  6,  respectively, 
and  whose  lateral  edge  is  5. 


PYRAMIDS. 


303 


41.  Find  the  lateral  area  and  volume  of  a  regular  quadrangular 
pyramid,  the  area  of  whose  base  is  100,  and  whose  lateral  edge  is  13. 

42.  The  lateral  surface  of  a  pyramid  is  greater  than  its  base. 
(From  foot  of  altitude  draw  lines  to  the  vertices  of  the  base  ;  each  A 

formed  has  a  smaller  altitude  than  the  corresponding  lateral  face.) 

43.  If  E,  F,  G,  and  H  are  the  middle  points  of  edges  AB,  AD, 
CD,  and  BC,  respectively,  of  tetraedron  A  BCD,  prove  EFGH  a 
parallelogram.     (§  130.) 

44.  Two  tetraedrons  are  equal  if  a  diedral  angle  and  the  adjacent 
faces  of  one  are  equal,  respectively,  to  a  diedral  angle  and  the  adjacent 
faces  of  the  other,  if  the  equal  parts  are  similarly  placed. 

(Figs,  of  §  459.  Given  faces  GAB,  OAC,  and  diedral  Z  OA  equal, 
respectively,  to  faces  O'A  B',  O'A'C,  and  diedral  Z  O'A'.) 

Dl. F' 

45.  The  section  of  a  prism  made  by  a  plane 
parallel  to  a  lateral  edge  is  a  parallelogram. 

(Given  section  EE'F'F  \\  AA  .     Prove  EE'  \\  to 
plane  CD' ;  then  use  §  412.) 


46.  The  point  of  intersection  of  the  diagonals 
of  a  parallelopiped  is  called  the  centre  of  the  par- 
allelopiped.     (Ex.  3.) 

Prove  that  any  line  drawn  through  the  centre 
of  a  parallelopiped,  terminating  in  a  pair  of  oppo- 
site faces,  is  bisected  at  that  point. 

47.  The  volume  of  a  regular  prism  is  equal  to  its  lateral  area, 
multiplied  by  one-half  the  apothem  of  its  base.     (§  350.) 

48.  The  volume  of  a  regular  pyramid  is  equal  to  its  lateral  area, 
multiplied  by  one-third  the  distance  from  the  centre  of  its  base  to  any 
lateral  face. 

(Pass  planes  through  the  lateral  edges  and  the  centre  of  the  base.) 

49.  Find  the  area  of  the  entire  surface  and  the  volume  of  a  trian- 
gular pyramid,  each  of  whose  edges  is  2. 

50.  The  areas  of  the  bases  of  a  frustum  of  a  pyramid  are  12  and 
75,  respectively,  and  its  altitude  is  9.  What  is  the  altitude  of  the 
pyramid  ? 

(Let  altitude  of  pyramid  =  x  ;  then  x  -  9  is  the  ±  from  its  vertex 
to  the  upper  base  of  the  frustum  ;  then  use  §  515.) 


304 


SOLID  GEOMETRY.— BOOK  VII. 


51.  The  bases  of  a  frustum  of  a  pyramid  are  rectangles,  whose  sides 
are  27  and  15,  and  9  and  5,  respectively,  and  the  line  joining  their 
centres  is  perpendicular  to  each  base.  If  the  altitude  of  the  frustum 
is  12,  find  its  lateral  area  and  volume. 

(From  the  centre  of  each  base  draw  Js  to  two  of  its  sides ;  in  this 
way  the  altitudes  of  the  lateral  faces  may  be  found.) 

52.  A  frustum  of  any  pyramid  is  equivalent  to  the  sum  of  three 
pyramids,  having  for  their  common  altitude  the  altitude  of  the  frus- 
tum, and  for  their  bases  the  lower  base,  the  upper  base,  and  a  mean 
proportional  between  the  bases,  of  the  frustum.     (§  524.) 

53.  The  upper  base  of  a  truncated  paral- 
lelopiped  is  a  parallelogram. 

(Let  planes  AC  and  BD'  intersect  in  00'  ;   ^^ 
prove  that  00'  bisects  A'C  and  B'D'.) 

54.  The  sum  of  two  opposite  lateral  edges  of  a  truncated  paral- 
lelopiped  is  equal  to  the  sum  of  the  other  two  lateral  edges. 

(Fig.  of  Ex.  53.  Find  the  length  of  00'  in  terms  of  the  lateral 
edges  by  §  132.) 


55.  The  volume  of  a  truncated  parallelo- 
piped  is  equal  to  the  area  of  a  right  section, 
multiplied  by  one-fourth  the  sum  of  the  lateral 


(By  proof  of  §  483,  a  rt.  section  of  a  paral- 
lelopiped  is  a  O ;  divide  the  solid  into  two 
truncated  triangular  prisms,  and  apply  Ex.  54.) 

56.  The  volume  of  a  truncated  parallelopiped  is  equal  to  the  area 
of  a  right  section,  multiplied  by  the  distance  between  the  centres  of 
the  bases. 

(By  Ex.  54,  the  distance  between  the  centres  of  the  bases  may  be 
proved  equal  to  one-fourth  the  sum  of  the  lateral  edges.) 

57.  If  ABCD  is  a  rectangle,  and  EF  any 
line  not  in  its  plane  parallel  to  AB^  the  vol- 
ume of  the  solid  bounded  by  figures  ABCD, 
ABFE,  CDEF,  ADE,  and  BGF,  is 

\hx  ADx  (2AB-hEF), 
where  h  is  the  perpendicular  from  any  point  of 
EF  to  ABCD.     (§  525.) 


PYRAMIDS. 


305 


58.  If  ABCD  and  EFGH  are  rectangles 
lying  in  parallel  planes,  AB  and  BC  being 
parallel  to  EF  and  FG,  respectively,  the  solid 
bounded  by  the  figures  ABCD,  EFGH,  ABFE, 
BCGF,  CDHG,  and  DAEH,  is  called  a  rec- 
tangular prismoid. 

ABCD  and  EFGH  are  called  the  bases  of 
the  rectangular  prismoid,  and  the  perpendicular 
distance  between  them  the  altitude. 

Prove  the  volume  of  a  rectangular  prismoid  equal  to  the  sum  of  its 
bases,  plus  four  times  a  section  equally  distant  from  the  bases,  multi- 
plied by  one-sixth  the  altitude. 

(Pass  a  plane  through  CD  and  EF,  and  find  volumes  of  solids 
ABCD-EF  and  EFGH-CD  by  Ex.  57.) 

59.  Find  the  volume  of  rectangular  prismoid  the  sides  of  whose 
bases  are  10  and  7,  and  6  and  5,  respectively,  and  whose  altitude  is  9. 

60.  Two  tetraedrons  are  equal  if  three  faces  of  one  are  equal,  re- 
spectively, to  three  faces  of  the  other,  if  the  equal  parts  are  similarly 
placed.     (§  460,  1.) 

61.  The  perpendicular  drawn  to  the  lower 
base  of  a  truncated  right  triangular  prism  from 
the  intersection  of  the  medians  of  the  upper 
base,  is  equal  to  one-third  the  sum  of  the 
lateral  edges. 

(Let  P  be  the  middle  point  of  DL,  and  draw 
PQLABC;  express  LM  in  terms  of  PQ  and 
6^iV^by  §  132.) 

62.  The  three  planes  passing  through  the  lateral  edges  of  a  tri- 
angular pyramid,  bisecting  the  sides  of  the  base,  meet  in  a  common 
straight  line. 

(Fig.  of  Ex.  24,  p.  272.  The  intersections  of  the  planes  with  the 
base  of  the  pyramid  are  the  medians  of  the  base.) 

63.  A  monument  is  in  the  form  of  a  frustum  of  a  regular  quad- 
rangular pyramid  8  ft.  in  height,  the  sides  of  whose  bases  are  3  ft.  and 
2  ft.,  respectively,  surmounted  by  a  regular  quadrangular  pyramid 
2  ft.  in  height,  each  side  of  whose  base  is  2  ft.  What  is  its  weight, 
at  180  lb.  to  the  cubic  foot  ? 

64.  Find  the  area  of  the  base  of  a  regular  quadrangular  pyramid, 
whose  lateral  faces  are  equilateral  triangles,  and  whose  altitude  is  5. 

(Represent  lateral  edge  and  side  of  base  by  x.) 


306 


SOLID   GEOMETRY.— BOOK  VII. 


65.    A  plane  passed  through  the  centre  of  a  parallelopiped  divides 
it  into  two  equivalent  solids.     (Ex.  55. ) 


66.  The  sides  of  the  base,  AB,  BC,  and  CA, 
of  truncated  right  triangular  prism  ABC-DEF 
are  15,  4,  and  12,  respectively,  and  the  lateral 
edges  AD,  BE,  and  CF  are  15,  7,  and  10,  re- 
spectively.    Find  the  area  of  upper  base  DEF. 

(Draw  EH±CF,  and  HG  and  FKJLAD. 
Find  area  DEF  by  §  324.) 


67.  The  volume  of  a  triangular  prism  is  equal  to  a  lateral  face, 
multiplied  by  one-half  its  perpendicular  distance  from  any  point  in 
the  opposite  lateral  edge. 

(Draw  a  rt.  section  of  the  prism,  and  apply  §  525.) 

68.  The  sum  of  the  squares  of  the  four 
diagonals  of  a  parallelopiped  is  equal  to  the 
sum  of  the  squares  of  its  twelve  edges. 

(To  prove  ACi^  +  AKI^  +  BD^  +  Wd^ 
equal  to  4  Zi^  +  4  Js"^  +  4  AD^  Apply 
Ex.  79,  p.  228,  to  OJAA'C'G.) 

A  '  B 

69.  The  altitude  and  lateral  edge  of  a  frustum  of  a  regular  tri- 
angular pyramid  are  8  and  10,  respectively,  and  each  side  of  its  upper 
base  is  2v^.     Find  its  volume  and  lateral  area. 


70.  If  ABCD  is  a  tetraedron,  the  section  made 
by  a  plane  parallel  to  each  of  the  edges  AB  and 
0Z>  is  a  parallelogram,     (§412.)  ^ 

(Toprove  J^iJ'G^lfaO.) 


71.  In  tetraedron  ABCD,  a  plane  is  drawn  through  edge  CD  per- 
pendicular to  AB,  intersecting  faces  ABC  and  ABD  in  CE  and  ED, 
respectively.     If  the  bisector  of  Z  CED  meets  CD  at  F,  prove 

CF:DF  =  area  ABC  :  area  ABD.     (§  249.) 

72.  The  sum  of  the  perpendiculars  drawn  to  the  faces  from  any 
point  within  a  regular  tetraedron  (§  536)  is  equal  to  its  altitude. 

(Divide  the  tetraedron  into  triangular  pyramids,  having  the  given 
point  for  their  common  vertex.) 


SIMILAR   POLYEDRONS. 


307 


73.   The  planes  bisecting  the  diedral  angles 
of  a  tetraedron  intersect  in  a  common  point. 


74.  If  the  four  diagonals  of  a  quadrangular  prism  pass  through  a 
common  point,  the  prism  is  a  parallelopiped. 

(In  Fig.  of  Ex.  68,  let  AC,  A'C,  BD\  and  B'D  pass  through  a 
common  point.    To  prove  J. C  a  parallelopiped.    Prove  ^C  a  O.) 


SIMILAR  POLYEDRONS. 

527.  Def.  Two  polyedrons  are  said  to  be  similar  when 
they  have  the  same  number  of  faces  similar  each  to  each 
and  similarly  placed,  and  have  their  homologous  polyedral 
angles  equal. 

Prop.  XXIV.     Theorem. 

528.  Tlie  ratio  of  any  two  homologous  edges  of  two  similar 
polyedrons  is  equal  to  the  ratio  of  any  other  two  homologous 
edges. 


Given,  in  similar  polyedrons  .AF  and  A'F',  edge  AB 
homologous  to  edge  A'B',  and  edge  EF  to  edge  E'F' ;  and 
faces  AO  and  I)F  similar  to  faces  A'C  and  D'F',  respec- 
tively. 

AB  ^  EF 
A'B'     E'F'' 
AB       CD 


To  Prove 


By  §  253,  2, 


A'B'     CD' 


308  SOLID  GEOMETRY.— BOOK   VII. 

529.  Cor.  I.  Any  two  homologous  faces  of  two  similar 
polyedrons  are  to  each  other  as  the  squares  of  any  two  homolo- 
gous edges. 

To  prove  ^^^!^,  =  MEr,    See  §  322.^ 
area  A'B'C'D'      E'F'  J 

530.  Cor.  II.  The  entire  surfaces  of  two  similar  polyedrons 
are  to  ea^h  other  as  the  squares  of  any  two  homologous  edges. 

/^To    rove      area  ABCD  +  area  CDEF  etc.      _  EF^  \ 
V      ^^  ^    area  A'B'C'D'  +  area  C'D'E'F'  etc.  ~  WF''' J 

Prop.  XXV.     Theorem. 

531.  Two  tetraedrons  are  similar  when  the  faces  including 
a  triedral  angle  of  one  are  similar,  respectively,  to  the  faces 
including  a  triedral  angle  of  the  other,  and  similarly  placed. 

A\ 


C 

Given,  in  tetraedrons  ABCD  and  A'B'C'D',  face  ABC 
similar  to  A'B'C,  ACD  to  A'C'D',  and  ADB  to  A'D'B'. 
To  Prove  ABCD  and  A'B'C'D'  similar. 
Proof.     From  the  given  similar  faces,  we  have 

B^^AG^CD_^AD^^BD^  m 

B'C     A'C     CD'     A'D'     B'D''  ^'^ 

Hence,  faces  BCD  and  B'C'D'  are  similar.  (§  259) 

Again,  A  BAG,  CAD,  and  DAB  are  equal,  respectively, 

to  AB'A'C,  C'A'D',  and  D'A'B'.  (?) 

Then,  triedral  AA-BCD  and  A'-B'C'D'  are  equal. 

(§  460,  1) 
Similarly,  any  two  homologous  triedral  A  are  equal. 
Therefore,  ABCD  and  A'B'C'D'  are  similar  (§  527). 


SIMILAR  POLYEDRONS.  3Q9 


Prop.  XXVI.     Theorem. 

532.  Two  tetraedrons  are  similar  when  a  diedral  angle  of 
one  is  equal  to  a  diedral  angle  of  the  other,  and  the  faces 
including  the  equal  diedral  angles  similar  each  to  each,  and 
similarly  placed. 


Given,  in  tetraedrons  ABGD  and  A'B'C'D\  diedral  Z  AB 
equal  to  diedral  Z  A'B' ;  and  faces  ABC  and  ABD  similar 
to  faces  A'B'C  and  A'B'D',  respectively. 

To  Prove  ABCD  and  A'B'C'D'  similar. 

Proof.  Apply  tetraedron  A'B'C'D'  to  ABCD  so  that  die- 
dral ZA'B'  shall  coincide  with  its  equal  diedral  ZAB, 
point  A'  falling  at  A. 

Then  since  Z  B'A'C  =  Z  BAG  and  Z  B'A'D'  =  Z  BAD, 
edge  A'C  will  coincide  with  edge  AC,  and  A'D'  with  AD. 
.'.  ZC'A'D'  =  ZCAD. 

Again,  from  the  given  similar  faces, 

ATP^AJB^^^A^D^  m 

AC       AB       ad'  ^'^ 

Hence,  A  C'A'D'  is  similar  to  A  CAD.  (§  261) 

Then,  the  faces  including  triedral  Z  A'-B'C'D'  are  similar 
respectively  to  the  faces  including  triedral  Z  A-BCD,  and 
similarly  placed. 

Therefore,  ABCD  and  A'B'C'D'  are  similar.  (§  531) 


Ex.  75.    If  a  tetpaedron  be  cut  by  a  plane  parallel  to  one  of  its 
faces,  the  tetraedron  cut  off  is  similar  to  the  givei^i  tetraedron. 


310  SOLID   GEOMETRY.— BOOK  VII. 


Prop.  XXVII.     Theorem. 

533.  Two  similar  polyedrons  may  be  decomposed  into  the 
same  number  of  tetraedrons,  similar  each  to  each,  and  simi- 
larly placed. 


Given  ^Fand  A^F^  similar  polyedrons,  vertices  JL  and  J.' 
being  homologous. 

To  Prove  that  they  may  be  decomposed  into  the  same 
number  of  tetraedrons,  similar  each  to  each,  and  similarly 
placed. 

Proof.  Divide  all  the  faces  of  AF,  except  the  ones  hav- 
ing ^  as  a  vertex,  into  A ;  and  draw  lines  from  A  to  their 
vertices. 

In  like  manner,  divide  all  the  faces  of  A^F',  except  the 
ones  having  A'  as  a  vertex,  into  A  similar  to  those  in  AF, 
and  similarly  placed.  (§  267) 

Draw  lines  from  A'  to  their  vertices. 

Then,  the  given  polyedrons  are  decomposed  into  the  same 
number  of  tetraedrons,  similarly  placed. 

Let  ABCF  2aidi  A'B'C'F'  be  homologous  tetraedrons. 

A  ABC  and  BCF  are  similar,  respectively,  to  AA'B'C 
and  B'C'F'.  (§  267) 

And  since  the  given  polyedrons  are  similar,  the  homolo- 
gous diedral  A  BC  and  B'C  are  equal. 

Therefore,  ABCF  and  A'B'C'F'  are  similar.  (§  532) 

In  like  manner,  we  may  prove  any  two  homologous 
tetraedrons  similar. 

Hence,  the  given  polyedrons  are  decomposed  into  the 
same  number  of  tetraedrons,  similar  each  to  each,  and 
similarly  placed. 


J 


SIMILAR  POLYEDRONS.  ^H 

Prop.  XXVIII.     Theorem. 

534.    Two  similar  tetraedrons  are  to  each  other  as  the  cubes 
of  their  homologous  edges. 


C 

Given  Fand  "F  the  volumes  of  similar  tetraedrons  ABCD 
and  A'B'C'V,  vertices  A  and  A'  being  homologous. 

To  Prove  Z^A^. 

Proof.     Since  the  triedral  A  2it  A  and  A'  are  equal, 
V        ABxACxAD 


V     A'B'xA'C'xA'D' 
AB  ^AC_^  AD 


523) 


A'B'     A'C     A'D' 


-p  .  AC       AB         -,    AD       AB  ..  ^oq\ 

^"*'  1^  =  1^'^"^  J^  =  ^-  ^^'''^ 

V  ^  AB       AB       AB  ^  A^ 
"   V     A'B'     A'B'     A'B'     jTb^' 

535.  Cor.  Any  two  similar  polyedrons  are  to  each  other  as 
the  cubes  of  their  homologous  edges. 

For  any  two  similar  polyedrons  may  be  decomposed 
into  the  same  number  of  tetraedrons,  similar  each  to  each 
(§  533). 

Any  two  homologous  tetraedrons  are  to  each  other  as  the 
cubes  of  their  homologous  edges.  (§  534) 

Then,  any  two  homologous  tetraedrons  are  to  each  other 
as  the  cubes  of  any  two  homologous  edges  of  the  polyedrons. 

(§  528) 


312  SOLID   GEOMETRY.  — BOOK   VII. 

REGULAR  POLYBDRONS. 

536.  Def.  A  regular  poly edron  is  a  polyedron  whose  faces 
are  equal  regular  polygons,  and  whose  polyedral  angles  are 
all  equal. 

Prop.  XXIX.     Theorem. 

537.  Not  more  than  Jive  regular  convex  polyedrons  are 
possible. 

A  convex  polyedral  Z  must  have  at  least  three  faces,  and 
the  sum  of  its  face  A  must  be  <  360°  (§  458). 

1.  With  equilateral  triangles. 

Since  the  Z  of  an  equilateral  A  is  60°,  we  may  form  a  con- 
vex polyedral  Z  by  combining  either  3,  4,  or  5  equilateral  A. 

Not  more  than  5  equilateral  A  can  be  combined  to  form  a 
convex  polyedral  Z.  (§  458) 

Hence,  not  more  than  three  regular  convex  polyedrons  can 
be  bounded  by  equilateral  A. 

2.  With  squares. 

Since  the  Z  of  a  square  is  90°,  we  may  form  a  convex 
polyedral  Z  by  combining  3  squares. 

Not  more  than  3  squares  can  be  combined  to  form  a  con- 
vex polyedral  Z.  (?) 

Hence,  not  more  than  one  regular  convex  polyedron  can 
be  bounded  by  squares. 

3.  With  regular  pentagons. 

Since  the  Z  of  a  regular  pentagon  is  108°,  we  may  form  a 
convex  polyedral  Z  by  combining  3  regular  pentagons. 

Not  more  than  3  regular  pentagons  can  be  combined  to 
form  a  convex  polyedral  Z.  (?) 

Hence,  not  more  than  one  regular  convex  polyedron  can 
be  bounded  by  regular  pentagons. 

Since  the  Z  of  a  regular  hexagon  is  120°,  no  convex  polye- 
dral Z  can  be  formed  by  combining  regular  hexagons.       (?) 


REGULAR  POLYEDRONS.  3;[3 

Hence,  no  regular  convex  polyedron  can  be  bounded  by 
regular  hexagons. 

In  like  manner,  no  regular  convex  polyedron  can  be 
bounded  by  regular  polygons  of  more  than  six  sides. 

Therefore,  not  more  than  five  regular  convex  polyedrons 
are  possible. 

Prop.  XXX.     Theorem. 
538.    With  a  given  edge,  to  construct  a  regular  polyedron. 

We  will  now  prove,  by  actual  construction,  that  five  regu- 
lar convex  polyedrons  are  possible : 

1.  The  regular  tetraedron,  bounded  by  4  equilateral  A. 

2.  The  regular  hexaedron,  or  cube,  bounded  by  6  squares. 

3.  The  regular  octaedron,  bounded  by  8  equilateral  A. 

4.  The  regular  dodecaedron,  bounded  by  12  regular  pen- 
tagons. 

5.  The  regular  icosaedron,  bounded  by  20  equilateral  A. 

1.    To  construct  a  regular  tetraedron. 
Given  line  AB. 

Required  to  construct  with  AB  as  an  a 

edge  a  regular  tetraedron.  / 

Construction.     Construct  the  equilateral         / 
A  ABC.  / 

At  its  centre  E,  draw  line  ED  J_  ABC',     ^^--^^ 
and  take  point  D  so  that  AD  =  AB.  ^^ 

Draw  lines  AD,  BD,  and  CD. 

Then,  solid  ABCD  is  a  regular  tetraedron. 

Proof.     Since  A,  B,  and  C  are  equally  distant  from  E, 

AD=:BD=  CD.  (§  406,  I) 

Hence,  the  six  edges  of  the  tetraedron  are  all  equal. 
Then,»the  faces  are  equal  equilateral  A.  (§  69) 

And  since  the  A  of  the  faces  are  all  equal,  the  triedral  A 
whose  vertices  are  A,  B,  C,  and  D  are  all  equal.     (§  460,  1) 
Therefore,  solid  ABCD  is  a  regular  tetraedron.       (§  536) 


314 


SOLID   GEOMETRY. —BOOK   VII. 


D^- 


(§  460,  1) 


2.  To  construct  a  regular  hexaedron,  or  ctibe. 
Given  line  AB. 
Required   to  construct   with  AB  as  an 

edge  a  cube.  e 

Construction.  Construct  square  ABCD ; 
and  draw  lines  AI],  BF,  CG,  and  DH,  each 
equal  to  AB,  and  _L  ABCD. 

Draw  lines  EF,  FG,  GH,  and  HE-,  then, 
solid  AG  is  a  cube. 

Proof.     By  cons.,  its  faces  are  equal  squares. 

Hence,  its  triedral  A  are  all  equal. 

3.  To  construct  a  regular  octaedron. 
Given  line  AB. 
Required  to  construct  with  AB  as  an 

edge  a  regular  octaedron. 

Construction.  Construct  the  square 
ABCD;  through  its  centre  0  draw  line 
EF±  ABCD,  making  OE=OF=  OA. 

Draw  lines  EA,  EB,  EC,  ED,  FA,  FB, 
FC,  and  FD ;  then  solid  AEFC  is  a  regular  octaedron. 

Proof.     Draw  lines  OA,  OB,  and  OD. 

Then  in  rt.  A  AOB,  AOE,  and  AOF,  by  cons., 
OA=OB=OE=  OF. 
.-.  A  AOB  =  A  AOE  =  A  AOF. 
.:  AB  =  AE  =  AF. 

Then,  the  eight  edges  terminating  at  E  and  F  are  all 
equal.  (§  406,  I) 

Thus,  the  twelve  edges  of  the  octaedron  are  all  equal,  and 
the  faces  are  equal  equilateral  A.  (?) 

Again,  by  cons.,  the  diagonals  of  quadrilateral  BEDF 
are  equal,  and  bisect  each  other  at  rt.  A. 

Hence,  BEDF  is  a  square  equal  to  ABCD,  and  OA  is  ± 
to  its  plane.  (§  400) 

Then,  pyramids  A-BEDF  and  E-ABCD  are  equal ;  and 
hence  polyedral  A  A-BEDF  and  E-ABCD  are  equal. 


(?) 

CO 


REGULAR  POLYEDRONS. 

In  like  manner,  any  two  polyedral  A  are  equal. 
Therefore,  solid  AEFC  is  a  regular  octaedron. 
4.    To  construct  a  regular  dodecaedron. 
d' 


315 


D 

Fig.  1. 

Given  line  AB. 

Required  to  construct  with  AB  as  an  edge  a  regular  do- 
decaedron. 

Construction.  Construct  regular  pentagon  ABODE  (Fig. 
1) ;  and  to  it  join  five  equal  regular  pentagons,  so  inclined  as 
to  form  equal  triedral  A  at   A,  B,  C,  D,  and  E.      (§  460,  1) 

Then  there  is  formed  a  convex  surface  AK  composed  of 
six  regular  pentagons,  as  shown  in  lower  part  of  Fig.  1. 

Construct  a  second  surface  A'K'  equal  to  AK,  as  shown 
in  upper  part  of  Fig.  1. 

Surfaces  AK  and  A'K'  may  be  combined  as  shown  in 
Fig.  2,  so  as  to  form  at  i^  a  triedral  Z  equal  to  that  at  A, 
having  for  its  faces  the  regular  pentagons  about  vertices  F 
and  F'  in  Fig.  1.  (§  460,  1) 

Then,  solid  AK  is  a  regular  dodecaedron. 

Proof.  Since  G'  falls  at  G,  and  diedral  Z  FG  and  face 
A  FGH  and  FGD'  (Fig.  2)  are  equal  respectively  to  the 
diedral  Z  and  face  A  of  triedral  A  F,  the  faces  about  vertex 
G  will  form  a  triedral  Z  equal  to  that  at  F. 

In  this  way,  it  may  be  proved  that  at  each  of  the  vertices 
H,  K,  etc.,  there  is  formed  a  triedral  Z  equal  to  that  at  F. 

Therefore,  solid  AKis  a  regular  dodecaedron. 


316 


SOLID   GEOMETRY.— BOOK  VII. 


5.    To  construct  a  regular  icosaedron. 

E-^ ^F 


Fig.  1. 

Given  line  AB. 


Fig.  3. 


Required  to  construct  with  AB  as  an  edge  a  regular 
icosaedron. 

Construction.  Construct  regular  pentagon  ABODE  (Fig. 
1) ;  at  its  centre  0  draw  line  OF  _L  ABODE,  making 
AF=  AB,  and  draw  lines  AF,  BF,  OF,  DF,  and  EF. 

Then,  F-ABODE  is  a  polyedral  Z  composed  of  five  equal 
equilateral  A.  (§§  406,  I,  69) 

Then  construct  two  other  polyedral  A,  A-BFEGH  and 
E-AFDKG,  each  equal  to  F-ABODE ;  and  place  them  as 
shown  in  upper  part  of  Fig.  2,  so  that  faces  ABF  and  AEF 
of  A-BFEGH,  and  faces  AEF  and  DEF  of  E-AFDKG, 
shall  coincide  with  the  corresponding  faces  of  F-ABODE. 

Then  there  is  formed  a  convex  surface  GO,  composed  of 
ten  equilateral  A. 

Construct  a  second  surface  (r'C  equal  to  GO,  as  shown 
in  lower  part  of  Fig.  2. 

Surfaces  GO  and  6r'C"  may  be  combined  as  shown  in 
Fig.  3,  so  that  edges  GH  and  HB  shall  coincide  with  edges 
G'H'  and  H'B\  respectively. 

Then,  solid  G^O  is  a  regular  icosaedron. 

Proof.  Since  diedral  A  AH,  E'H',  and  F'H'  are  equal  to 
the  diedral  A  of  polyedral  Z  F,  the  faces  about  vertices 
Hand  H'  form  a  polyedral  Z  at  iJ equal  to  that  at  F. 


REGULAR   POLYEDRONS. 


817 


Then,  since  diedral  A  FB,  AB,  HB,  and  F'B  (Fig.  3)  are 
equal  to  the  diedral  A  of  polyedral  AF,  the  faces  about 
vertex  B  form  a  polyedral  A  equal  to  that  at  F;  and  it  may 
be  shown  that  at  each  of  the  vertices  C,  D,  etc.,  there  is 
formed  a  polyedral  A  equal  to  that  at  F. 

Therefore,  solid  OC  is  a  regular  icosaedron. 

539.  Sch.  To  construct  the  regular  polyedrons,  draw  the 
following  figures  on  cardboard ;  cut  them  out  entire,  and  on 
the  interior  lines  cut  the  cardboard  half  through ;  the  edges 
may  then  be  brought  together  to  form  the  respective  solids. 


Tetraedron. 


Hexaedron. 


OCTAEDRON. 


DODECAEDRON. 


Icosaedron. 


EXERCISES. 

76.    The  volume  of  a  pyramid  whose  altitude  is  7  in,  is  686  cu.  in. 
Find  the  volume  of  a  similar  pyramid  whose  altitude  is  12  in. 


318  SOLID   GEOMETRY.— BOOK   VII. 

77.  If  the  volume  of  a  prism  whose  altitude  is  9  ft.  is  171  cu.  ft., 
find  the  altitude  of  a  similar  prism  whose  volume  is  50f  cu.  ft. 

(Represent  the  altitude  by  x.) 

78.  Two  bins  of  similar  form  contain,  respectively,  375  and  648 
bushels  of  wheat.  If  the  first  bin  is  3  ft.  9  in.  long,  what  is  the  length 
of  the  second  ? 

79.  A  pyramid  whose  altitude  is  10  in.,  weighs  24  lb.  At  what 
distance  from  its  vertex  must  it  be  cut  by  a  plane  parallel  to  its  base 
so  that  the  frustum  cut  off  may  weigh  12  lb.  ? 

80.  An  edge  of  a  polyedron  is  56,  and  the  homologous  edge  of  a 
similar  polyedron  is  21.  The  area  of  the  entire  surface  of  the  second 
polyedron  is  135,  and  its  volume  is  162.  Find  the  area  of  the  entire 
surface,  and  the  volume,  of  the  first  polyedron. 

81.  The  area  of  the  entire  surface  of  a  tetraedron  is  147,  and  its 
volume  is  686.  If  the  area  of  the  entire  surface  of  a  similar  tetrae- 
dron is  48,  what  is  its  volume  ? 

(Let  X  and  y  denote  the  homologous  edges  of  the  tetraedrons. ) 

82.  The  area  of  the  entire  surface  of  a  tetraedron  is  75,  and  its 
volume  is  500.  If  the  volume  of  a  similar  tetraedron  is  32,  what  is 
the  area  of  its  entire  surface  ? 

83.  The  homologous  edges  of  three  similar  tetraedrons  are  3,  4, 
and  5,  respectively.  Eind  the  homologous  edge  of  a  similar  tetrae- 
dron equivalent  to  their  sum. 

(Represent  the  edge  by  x.) 

84.  State  and  prove  the  converse  of  Prop.  XXVII. 

85.  The  volume  of  a  regular  tetraedron  is  equal  to  the  cube  of  its 
edge  multiplied  by  jV^2. 

86.  The  volume  of  a  regular  tetraedron  is  18  V2.  Find  the  area  of 
its  entire  surface.     (Ex.  85.) 

(Represent  the  edge  by  x.) 

87.  The  volume  of  a  regular  octaedron  is  equal  to  the  cube  of  its 
edge  multiplied  by  |V2. 


Book  VIII. 


THE  CYLINDER,  CONE,  AND  SPHERE. 
DEFINITIONS. 

540.   A  cylindrical  surface  is  a  surface  generated  by  a 
moving  straight   line,  which  constantly  intersects  a  given 
plane  curve,  and  in  all  its  positions 
is  parallel  to  a  given  straight  line, 
not  in  the  plane  of  the  curve. 

Thus,  if  line  AB  moves  so  as  to 
constantly  intersect  plane  curve 
AD,  and  is  constantly  parallel  to 
line  MN,  not  in  the  plane  of  the 
curve,  it  generates  a  cylindrical 
surface. 

The  moving  line  is  called  the  generatrix,  and  the  curve 
the  directrix. 

Any  position  of  the  generatrix,  as  EF,  is  called  an  element 
of  the  surface. 

A  cylinder  is  a  solid  bounded  by  a  cylin- 
drical surface,  and  two  parallel  planes. 

The  parallel  planes  are  called  the  bases 
of  the  cylinder,  and  the  cylindrical  surface 
the  lateral  surface. 

The  altitude  of  a  cylinder  is  the  perpen- 
dicular distance  between  the  planes  of  its 
bases. 

A  light  cylinder  is  a  cylinder  the  elements  of  whose  lateral 
surface  are  perpendicular  to  its  bases. 

A  circular  cylinder  is  a  cylinder  whose  base  is  a  circle. 
319 


320  SOLID   GEOMETRY.  — BOOK   VIII. 

A  plane  is  said  to  be  tangent  to  a  cylinder  when  it  con- 
tains one,  and  only  one,  element  of  the  lateral  surface. 

541.  It  follows  from  the  definition  of  a  cylinder  (§  540) 
that 

The  elements  of  the  lateral  surface  of  a  cylinder  are  equal 
and  parallel.  (§  415) 

Prop.  I.     Theorem. 

542.  A  section  of  a   cylinder  made  by  a  plane  passing 
through  an  element  of  the  lateral  surface  is  a  parallelogram. 


Given  ABCD  a  section  of  cylinder  AF,  made  by  a  plane 
passing  through  AB,  an  element  of  the  lateral  surface. 

To  Prove  section  ABCD  a  O. 

Note.  It  should  be  observed  that,  with  the  above  hypothesis,  CD 
simply  represents  the  intersection  of  plane  AC  with  the  cylindrical 
surface,  and  may  be  a  curved  line  ;  it  must  be  proved  that  it  is  a  str. 
line  II  AB. 

Proof.     AD  and  BC  are  str.  lines,  and  II.        (§§  396,  414) 

Now  draw  str.  line  CE  in  plane  AC  II  AB;  then,  CE  is  an 
element  of  the  cylindrical  surface.  (§§  541,  53) 

Then  since  CE  lies  in  plane  AC,  and  also  in  the  cylin- 
drical surface,  it  must  be  the  intersection  of  the  plane  with 
the  cylindrical  surface. 

Then,  CD  is  a  str.  line  II  AB,  and  ABCD  is  a  O. 

543.  Cor.  A  section  of  a  right  cylinder  made  by  a  plane 
perpendicular  to  its  base  is  a  rectangle. 


THE   CYLINDER. 

Prop.  II.     Theorem. 
544.    The  bases  of  a  cylinder  are  equal. 

^ Q' 


321 


Given  cylinder  AB'. 

To  Prove  base  A'B'  =  base  AB. 

Proof.  Let  E',  F,  and  G'  be  any  three  points  in  the  perim- 
eter of  base  A^B\  and  draw  EE',  FF',  and  GG'  elements 
of  the  lateral  surface. 

Draw  lines  EF,  FG,  GE,  E'F',  F'G',  and  G'E'. 

Now,  EE'  and  FF'  are  equal  and  II.  (§  541) 

Then,  EE'F'F  is  a  O.  (?) 

.-.  E'F'  =  EF.  (?) 

Similarly,  E'G'  =  EG  and  F'G'  =  FG. 

.'.  AE'F'G'  =  AEFG.  (?) 

Then,  base  A'B'  may  be  superposed  upon  base  AB  so  that 
points  E',  F',  and  G'  shall  fall  at  E,  F,  and  G,  respectively. 

But  E'  is  any  point  in  the  perimeter  of  A'B'. 

Then,  every  point  in  the  perimeter  of  A'B'  will  fall  some- 
where in  the  perimeter  of  AB,  and  base  A'B'  =  base  AB. 

545.  Cor.  I.  The  sections  of  a  circular  cylinder  made  by 
planes  parallel  to  its  bases  are  equal  circles. 

For  each  may  be  regarded  as  the  upper  base  of  a  cylinder 
whose  lower  base  is  a  O. 

546.  Def.  The  axis  of  a  circular  cylinder  is  a  straight 
line  drawn  between  the  centres  of  its  bases. 


322 


SOLID   GEOMETRY. —BOOK  VIII. 


547.  Cor.  II.  The  axis  of  a  circular  cylinder  is  parallel 
to  the  elements  of  its  lateral  surface. 

Given  AA'  the  axis,  and  BB'  an  ele- 
ment of  the  lateral  surface,  of  circular 
cylinder  BC 

To  Prove        AA'  II  BB'. 

Proof.     Let  BB'C'G  be  a  section  made 

by  a  plane  passing  through  BB'  and  A ; 

then5i5'(7'(7isaO.  (§542) 

.-.  B'C'  =  Ba  (?) 

Then  since  BC  is  a  diameter  of  O  BC,  and  ©  BC  and 
B'C  are  equal,  B'C  is  a  diameter  of  QB'C,  and  passes 
through  A'. 

Hence,  AB  and  ^'5'  are  equal  and  II.  (?) 

Then,  ABB' A'  is  a  O.  (?) 

.-.  AA'  II  55'. 

54B.  Cor.  III.  The  axis  of  a  circular  cylinder  passes 
through  the  centres  of  all  sections  parallel  to  the  bases. 


Prop.  III.     Theorem. 

549.   A  right  circular  cylinder  may  he  generated  by  the 
revolution  of  a  rectangle  about  one  of  its  sides  as  an  axis. 


Given  rect.  ABCD. 

To  Prove  the  solid  generated  by  the  revolution  of  ABCD 
about  AD  as  an  axis  a  rt.  circular  cylinder. 
Proof.     All  positions  of  BC  are  II  AD. 
Again,  AB  and  CD  generate  CD  ±  AD.  (§  402) 

Then,  these  (D  are  II,  and  ±  BC.  (§§  421,  419) 

Whence,  ABCD  generates  a  rt.  circular  cylinder. 


THE   CYLINDER.  323 

550.  Defs,  From  the  property  proved  in  §  549,  a  right 
circular  cylinder  is  called  a  cylinder  of  revolution. 

Similar  cylinders  of  revolution  are  cylinders  generated 
by  the  revolution  of  similar  rectangles  about  homologous 
sides  as  axes. 

Prop.  IV.     Theorem. 

551.  A  plane  drawn  through  an  element  of  the  lateral  sur- 
face of  a  circular  cylinder  and  a  tangent  to  the  base  at  its 
extremity,  is  tangent  to  the  cylinder. 


Given  AA^  an  element  of  the  lateral  surface  of  circular 
cylinder  AB\  line  CD  tangent  to  base  AB  at  A,  and  plane 
CD'  drawn  through  AA!  and  CD. 

To  Prove  CD^  tangent  to  the  cylinder. 

Proof.  Let  E  be  any  point  in  plane  CD\  not  in  AA\ 
and  draw  through  E  a  plane  II  to  the  bases,  intersecting  CD' 
in  line  EF,  and  the  cylinder  in  O  FH.  (§  545) 

Draw  axis  00' ;  then  00'  is  II  AA'.  (§  547) 

Let  the  plane  of  00'  and  AA'  intersect  the  planes  of  AB 
and  FH  in  radii  OA  and  GF,  respectively.  (§  548) 

Then,  OF  II  OA  and  FE  II  AD.  (§  414) 

.-.  ZGFE  =  ZOAD.  (§426) 

But  Z  OAD  is  a  rt.  Z.  (§  1'^^) 

Then,  FE  is  ±  GF,  and  tangent  to  O  FH.  (§  169) 

Whence,  point  E  lies  without  the  cylinder. 

Then,  all  portions  of  CD',  not  in  AA',  lie  without  the 
cylinder,  and  CD'  is  tangent  to  the  cylinder. 


324  SOLID   GEOMETUV.  —  BOOK   VIII. 

552.    Cor.     A  plane  tangent  to  a  circular  cylinder  intersects 
the  planes  of  the  bases  in  lines  which  are  tangent  to  the  bases. 


Ex.  1.  The  sections  of  a  cylinder  made  by  two  parallel  planes 
which  cut  all  the  elements  of  its  lateral  surface  are  equal. 

THE  CONE. 
DEFINITIONS. 

553.  A  conical  surface  is  a  surface  generated  by  a  moving 
straight  line,  which  constantly  intersects 

a  given  plane  curve,  and  passes  through  M- --. 

a  given  point  not  in  the  plane  of  the  ^  vi      ^^^ 

curve.  \\    y^ 

Thus,  if  line  OA  moves  so  as  to  con-  U^ 

stantly  intersect  plane  curve  ABC,  and  y"^  \\ 

constantly  passes  through  point  0,  not    a/^ V" \c 

in  the  plane  of  the  curve,  it  generates  a       n \y 

conical  surface.  ^ 

The  moving  line  is  called  the  generatrix,  and  the  curve 
the  directrix. 

The  given  point  is  called  the  vertex,  and  any  position  of 
the  generatrix,  as  OB,  is  called  an  element  of  the  surface. 

If  the  generatrix  be  supposed  indefinite  in  length,  it  will 
generate  two  conical  surfaces  of  indefinite  extent,  O-A'B'C 
and  0-ABC. 

These  are  called  the  upper  and  lower  nappes,  respectively. 

A  co7ie  is  a  solid  bounded  by  a  conical  surface,  and  a 
plane  cutting  all  its  elements. 

The  plane  is  called  the  base  of  the  cone,  and  /K 

the  conical  surface  the  lateral  surface.  /    \  \ 

The  altitude  of  a  cone  is  the  perpendicular       / [  \ 

distance  from  the  vertex  to  the  plane  of  the    /            1   A 
base.  ^-^ -^ 

A  circular   cone  is  a  cone  whose  base  is  a  circle. 


THE   CONE.  325 

The  axis  of  a  circular  cone  is  a  straight  line  drawn  from 
the  vertex  to  the  centre  of  the  base. 

A  right  circular  cone  is  a  circular  cone  whose  axis  is  per- 
pendicular to  its  base. 

A  frustum  of  a  cone  is  a  portion   of  a  cone   included 
between  the  base  and  a  plane  parallel  to 
the  base.  /^        ^ 

The  base  of  the  cone  is  called  the  loicer         /  ( 

base,  and  the  section  made  by  the  plane  the      /- -,^   j 

upper  base,  of  the  frustum.  (  V 

The  altitude  is  the  perpendicular  distance     \. ^ 

between  the  planes  of  the  bases. 

A  plane  is  said  to  be  tangent  to  a  cone,  or  frustum  of  a 
cone,  when  it  contains  one,  and  only  one,  element  of  the 
lateral  surface. 

Prop.  V.     Theorem. 

554.  A  right  circular  cone  may  be  generated  by  the  revo- 
lution of  a  right  triangle  about  one  of  its  legs  as  an  axis. 


Given  C  the  rt.  Z  of  rt.  A  ABC. 

To  Prove  the  solid  generated  by  the  revolution  of  ABC 
about  AC  as  an  axis  a  right  circular  cone. 
(The  proof  is  left  to  the  pupil.) 

555.  Defs.  "From  the  above  property,  a  right  circular 
cone  is  called  a  cone  of  revolution. 

Similar  cones  of  revolution  are  cones  generated  by  the 
revolution  of  similar  right  triangles  about  homologous  legs 
as  axes. 


326  SOLID   GEOMETRY.— BOOK  VIII. 


Prop.  VI.     Theorem. 

556.   A  section  of  a  cone  made  by  a  plane  passing  through 
the  vertex  is  a  triangle. 


Given  OCD  a  section  of  cone  GAB  made  by  a  plane  pass- 
ing through  vertex  0. 

To  Prove  section  OCD  a  A. 

Proof.     We  have  CD  a  str.  line.  (§  396) 

Now  draw  str.  lines  in  plane  OCD  from  0  to  C  and  D ; 
these  str.  lines  are  elements  of  the  conical  surface.     (§  550) 

Then,  since  these  str.  lines  lie  in  plane  OCD,  and  also 
in  the  conical  surface,  they  must  be  the  intersections  of  the 
plane  with  the  conical  surface. 

Then,  OC  and  OD  are  str.  lines,  and  OCD  is  a  A. 

Prop.  VII.     Theorem. 

557.  A  section  of  a  circular  cone  made  by  a  plane  parallel 
to  the  base  is  a  circle. 


Given  A'B'O  a  section  of  circular  cone  S-ABC,  made  by 
a  plane  II  to  the  base. 
To  Prove  A'B'C  a  O. 


THE   CONE. 


327 


Proof.     Draw  axis  OS,  intersecting  plane  xl'B'C  at  0'. 
Let  A'  and  B'  be  any  two  points  in  perimeter  A'B'C. 
Let  the  planes  determined  by  these  points  and  OS  inter- 
sect the  base  in  radii  OA  and  OB,  the  section  in  lines  O'A' 
and  O'B',  and  the  lateral  surface  in  lines  SA'A  and  SB'B, 
respectively. 

Then,  SA'A  and  SB'B  are  str.  lines.  (§  556) 

Now,  O'A'  II  OA  and  O'B'  11  OB.  (§  414) 

Then,  A  SO' A'  and   SO'B'   are  similar  to  A.S'ok  and 


SOB,  respectively. 

O'A' 


OA 


^^'  and  ^'^' 


OB 
O'B' 

ob' 


SO' 

so' 


(§  257) 
(?) 


(?) 
(§  143) 


OA 

But,  0^  =  OB. 

Then,  0'^'  =  0'5' ;  and  as  A'  and  B'  are  aw?/  two  points 
in  perimeter  A'B'C,  section  A'B'C  is  sl  O. 

558.  Cor.  TJie  axis  of  a  circular  cone  passes  through  the 
centre  of  every  section  parallel  to  the  base. 

Prop.  YIII.     Theorem. 

559.  A  plane  drawn  through  an  element  of  the  lateral  sur- 
face of  a  circular  cone  and  a  tangent  to  the  base  at  its  extremity, 
is  tangent  to  the  cone.  .o 


Given  OA  an  element  of  the  lateral  surface  of  circular 
cone  OAB,  line  CD  tangent  to  base  AB  at  A,  and  plane 
OCB  drawn  through  OA  and  OD. 


328  SOLID   GEOMETRY.  —  BOOK  VIIL 

To  Prove  OCD  tangent  to  the  cone. 
(Prove  that  E  lies  without  the  cone.) 

560.  Cor.  A  plane  tangent  to  a  circular  cone  intersects  the 
plane  of  the  base  in  a  line  tangent  to  the  base. 

THE  SPHERE. 
DEFINITIONS. 

561.  A  sphere  is  a  solid  bounded  by  a  surface,  all  points 
of  which  are  equally  distant  from  a  point  within  called  the 
centre. 

A  radius  of  a  sphere  is  a  straight  line  drawn  from  the 
centre  to  the  surface. 

A  diameter  is  a  straight  line  drawn  through  the  centre, 
having  its  extremities  in  the  surface. 

562.  It  follows  from  the  definition  of  §  561  that  all  radii 
of  a  sphere  are  equal. 

Also,  all  its  diameters  are  equal,  since  each  is  the  sum  of 
two  radii. 

563.  Two  spheres  are  equal  when  their  radii  are  equal. 
For  they  can  evidently  be  applied  one  to  the  other  so 

that  their  surfaces  shall  coincide  throughout. 
Conversely,  the  radii  of  equal  spheres  are  equal. 

564.  A  line  (or  a  plane)  is  said  to  be  tangent  to  a  sphere 
when  it  has  one,  and  only  one,  point  in  common  with  the 
surface ;  the  common  point  is  called  the  point  of  contact. 

A  polyedron  is  said  to  be  inscribed  in  a  sphere  when  all 
its  vertices  lie  in  the  surface  of  the  sphere ;  in  this  case  the 
sphere  is  said  to  be  circumscribed  about  the  polyedron. 

A  polyedron  is  said  to  be  circumscribed  about  a  sphere 
when  all  its  faces  are  tangent  to  the  sphere;  in  this  case 
the  sphere  is  said  to  be  inscribed  in  the  polyedron. 

565.  A  sphere  may  be  generated  by  the  revolution  of  a  semi- 
circle about  its  diameter  as  an  axis. 


THE   SPHERE. 


329 


For  all  points  of  such  a  surface  are  equally  distant  from 
the  centre  of  the  O.  (?) 

Prop.  IX.     Theorem. 
566.   A  section  of  a  sphere  made  by  a  plane  is  a  circle. 


6 


Given  ABC  a  section  of  sphere  AFC  made  by  a  plane. 

To  Prove  ABC  a  O. 

Proof.  Let  0  be  the  centre  of  the  sphere,  and  draw  line 
00'  ±  to  plane  ABC. 

Let  A  and  B  be  any  two  points  in  perimeter  ABC,  and 
draw  lines  OA,  OB,  OA,  and  O'B. 

Now,  •  OA=OB.  (?) 

.-.  OA=0'B.  (§407,1) 

But  A  and  B  are  any  two  points  in  perimeter  ABC. 

Therefore,  ABC  is  a  O. 

567.  Defs.     A  great  circle  of  a  sphere  is  a  section  made 
by  a  plane  passing  through  the  centre; 
as  ABC 

A  small  circle  is  a  section  made  by  a 
plane  which  does  not  pass  through  the 
centre. 

The  diameter  perpendicular  to  a  circle 
of  a  sphere  is  called  the  axis  of  the  circle, 
and  its  extremities  are  called  the  poles. 

568.  Cor.  I.  TJie  axis  of  a  circle  of  a  sphere  passes  through 
the  centre  of  the  circle. 


330  SOLID   GEOMETRY.— BOOK  VIII. 

569.  Cor.  II.     All  great  circles  of  a  sphere  are  equal. 
For  their  radii  are  radii  of  the  sphere. 

570.  Cor.  III.  Every  great  circle  bisects  the  sphere  and  its 
surface. 

For  if  the  portions  of  the  sphere  formed  by  the  plane  of 
the  great  O  be  separated,  and  placed  so  that  their  plane  sur- 
faces coincide,  the  spherical  surfaces  falling  on  the  same 
side  of  this  plane,  the  two  spherical  surfaces  will  coincide 
throughout ;  for  all  points  of  either  surface  are  equally  dis- 
tant from  the  centre. 

571.  Cor.  rV.     Any  two  great  circles  bisect  each  other. 

For  the  intersection  of  their  planes  is  a  diameter  of  the 
sphere,  and  therefore  a  diameter  of  each  O.  (§  152) 

572.  Cor.  V.  Between  any  two  points  on  the  surface  of  a 
sphere,  not  the  extremities  of  a  diameter,  an  arc  of  a  great 
circle,  less  than  a  semi-circumference,  can  be  drawn,  and  but 
one. 

For  the  two  points,  with  the  centre  of  the  sphere,  deter- 
mine a  plane  which  intersects  the  surface  of  the  sphere  in 
the  required  arc. 

Note.  If  the  points  are  the  extremities  of  a  diameter,  an  indefi- 
nitely great  number  of  arcs  of  great  ©  can  be  drawn  between  them  ; 
for  an  indefinitely  great  number  of  planes  can  be  drawn  through  the 
diameter. 

573.  Def.  The  distance  between  two 
points  on  the  surface  of  a  sphere,  not  at 
the  extremities  of  a  diameter,  is  the  arc  of 
a  great  circle,  less  than  a  semi-circum- 
ference, drawn  between  them. 

Thus,  the  distance  between  points  C  and 
D  is  arc  CED,  and  not  arc  CAFBD. 

574.  Cor.  VI.  An  arc  of  a  circle  may  be  drawn  through 
any  three  points  on  the  surface  of  a  sphere. 

For  the  three  points  determine  a  plane  which  intersects 
the  surface  of  the  sphere  in  the  required  arc. 


THE   SPHERE.  33^ 


Prop.  X.     Theorem. 

575.   All  points  in  the  circumference  of  a  circle  of  a  sphere 
are  equally  distant  from  each  of  its  poles. 


Given  P  and  P'  the  poles  of  O  ABC  of  sphere  APC. 

To  Prove  all  points  in  circumference  ABC  equally  distant 
(§  573)  from  P,  and  also  from  P'. 

Proof.     Let  A  and  B  be  any  two  points  in  circumference 
ABC,  and  draw  arcs  of  great  ©  PA  and  PB. 
Draw  axis  PP\  intersecting  plane  ABC  at  0. 
Draw  lines  OA  and  OB,  and  chords  PA  and  PB. 
Now  0  is  the  centre  of  O  ABC  (§  568) 

.-.  0A=  OB.  (?) 

.-.  chord  PA  =  chord  PB.  (§  406,  I) 

.-.  arcP^  =  arcP^.  (§  157) 

But  A  and  B  are  any  two  points  in  circumference  ABC. 

Therefore,  all  points  in  circumference  ABC  are  equally 
distant  from  P. 

In  like  manner,  all  points  in  circumference  ABC  are 
equally  distant  from  P'. 

576.  Def.  The  polar  distance  of  a  circle  of  a  sphere  is  the 
distance  (§  573)  from  the  nearer  of  its  poles,  or  from  either 
pole  if  they  are  equally  near,  to  the  circumference. 

Thus,  in  figure  of  Prop.  X,  the  polar  distance  of  O  ABC 
is  arc  PA. 


332 


SOLID   GEOMETRY.— BOOK  VIII. 


577.  Cor.  All  points  in  the  circumfer- 
ence of  a  great  circle  of  a  sphere  are  at  a 
quadranfs  distance  from  either  pole. 

Given  P  a  pole  of  great  Q  ABC  of 
sphere  APC,  B  any  point  in  circumfer- 
ence ABC,  and  PB  an  arc  of  a  great  O. 

To  Prove  arc  PB  a  quadrant  (§  146). 

Proof.     Let  0  be  the  centre  of  the  sphere,  and  draw  radii 
OB  and  OP. 
Then,  Z  POB  is  a  rt.  Z.  (§  398) 

Whence,  arc  PB  is  a  quadrant.  (§  191) 

The  above  proof  holds  for  either  pole  of  the  great  O. 

Note.  An  arc  of  a  circle  may  be  drawn  on  the  surface  of  a  sphere 
by  placing  one  foot  of  the  compasses  at  the  nearer  pole  of  the  circle, 
the  distance  between  the  feet  being  equal  to  the  chord  of  the  polar 
distance. 

Prop.  XI.     Theorem. 

578.  If  a  point  on  the  surface  of  a  sphere  lies  at  a  quad- 
rants distance  from  each  of  two  points  in  the  arc  of  a  great 
cii'cle,  it  is  a  pole  of  that  arc. 

Note.  The  term  quadrant^  in  Spherical  Geometry,  usually  signi- 
fies a  quadrant  of  a  great  circle. 

P 


Given  point  P  on  surface  of  sphere  APC,  AB  an  arc  of 
great  O  ABC,  and  PA  and  PB  quadrants. 

To  Prove  P  a  pole  of  arc  AB. 

(PO  is  ±  to  OA  and  OB;  then  use  §  400.) 


THE    SPHERE. 


333 


Prop.  XII.     Theorem. 

579.  The  intersection  of  two  spheres  is  a  circle,  whose 
centre  is  in  the  straight  line  joining  the  centres  of  the  spheres, 
and  whose  plane  is  perpendicular  to  that  line. 


Given  two  intersecting  spheres. 

To  Prove  their  intersection  a  O,  whose  centre  is  in  the 
line  joining  the  centres  of  the  spheres,  and  whose  plane  is 
±  to  this  line. 

Proof.  Let  0  and  0'  be  the  centres  of  two  ©,  whose 
common  chord  is  AB;  draw  line  00',  intersecting  ^5  at  C. 

Then,  00'  bisects  AB  at  rt.  A.  (§  178) 

If  we  revolve  the  entire  figure  about  00'  as  an  axis,  the 
(D  will  generate  spheres  whose  centres  are  0  and  0'.   (§  565) 

And  AC  will  generate  a  O  ±  00',  whose  centre  is  C, 
which  is  the  intersection  of  the  two  spheres.  (§  402) 


Prop.  XIII.     Theorem. 

580.   A  plane  perpendicular  to  a  radius  of  a  sphere  at  its 
extremity  is  tangent  to  the  sphere. 

C 


(The  proof  is  left  to  the  pupil ;  compare  §  169.) 


334  SOLID  GEOMETRY.— BOOK   VIII. 

581.  Cor.  (Converse  of  Prop.  XIII.)  A  j)lane  tangent  to 
a  sphere  is  perpendicular'  to  the  radius  drawn  to  the  2)oint  of 
contact.     (Fig.  of  Prop.  XIII.) 

(The  proof  is  left  to  the  pupil ;  compare  §  170.) 

Prop.  XIV.     Theorem. 

582.  Through  four  poi7its,  not  in  the  same  plane,  a  spherical 
surface  can  he  made  to  pass,  and  hut  one. 

A 


C 

Given  A,  B,  C,  and  D  points  not  in  the  same  plane. 

To  Prove  that  a  spherical  surface  can  be  passed  through 
A,  B,  C,  and  D,  and  but  one. 

Proof.  Pass  planes  through  A,  B,  C,  and  D,  forming 
tetraedron  ABCD,  and  let  K  be  the  middle  point  of  CD. 

Draw  lines  KE  and  KF  in  faces  ACD  and  BCD,  respec- 
tively, ±  CD ;  and  let  E  and  F  be  the  centres  of  the  cir- 
cumscribed ©  of  A  ACD  and  BCD,  respectively.        (§  222) 

Then  plane  EKF  is  ±  CD.  (§  400) 

Draw  line  EG  1.  ACD,  and  line  FH1.BCD;  then  EG 
and  FH  lie  in  plane  EKF.  (§  439). 

Then  EG  and  FH  must  meet  at  some  point  0,  unless 
they  are  II ;  this  cannot  be  unless  ACD  and  BCD  are  in  the 
same  plane,  which  is  contrary  to  the  hyp.  (§  418) 

Now  0,  being  in  EG,  is  equally  distant  from  A,  C,  and  D; 
and  being  in  FH,  is  equally  distant  from  B,  C,  and  D. 

(§  406,  I) 

Then  0  is  equally  distant  from  A,  B,  C,  and  D ;  and  a 
spherical  surface  described  with  0  as  a  centre,  and  OA  as  a 
radius,  will  pass  through  A,  B,  C,  and  Z). 


THE   SPHERE.  ggc 

Now  the  centre  of  any  spherical  surface  passing  through 
A,  B,  C,  and  D  must  be  in  each  of  the  Js  EG  and  FH. 

Then  as  EG  and  FH  intersect  in  but  one  point,  only  one 
spherical  surface  can  be  passed  through  A,  B,  C,  and  D. 

583.  Defs.  The  angle  between  two  intersecting  curves  is 
the  angle  between  tangents  to  the  curves  at  their  point  of 
intersection. 

A  spherical  angle  is  the  angle  between  two  intersecting 
arcs  of  great  circles. 

Prop.  XV.     Theorem. 

584.  A  spherical  ayigle  is  measured  by  an  arc  of  a  great 
circle  having  its  vertex  as  a  pole,  included  betioeen  its  sides 
produced  if  necessary. 

A 


C 

Given  ABC  and  AB'C  arcs  of  great  (D  on  the  surface  of 
sphere  AC,  lines  AD  and  AD'  tangent  to  ABC  and  A'BC, 
respectively,  and  BB'  an  arc  of  a  great  0  having  ^  as  a 
pole,  included  between  arcs  ABC  and  AB'C. 

To  Prove  that  A  DAD'  is  measured  by  arc  BB'. 

Proof.  Let  0  be  the  centre  of  the  sphere,  and  draw 
diameter  AOC  and  lines  OB  and  OB'. 

Now,  arcs  AB  and  AB'  are  quadrants.  (§  577) 

Whence,  z^^05  and  ^05' are  rt.  Zs.  (?) 

Therefore,  OB  II  AD  and  OB'  II  AD'.  (§§  170,  54) 

.-.  ZDAD'  =  ZBOB'.  (§426) 

But  Z  BOB'  is  measured  by  arc  BB'.  (?) 

Then,  Z  DAD'  is  measured  by  arc  BB'. 


336  SOLID   GEOMETRY. —BOOK  VIII. 

585.  Cor.  I.    (Fig.  of  Prop.  XV.)    Plane  BOB'  is  ±  OA. 

(§  400) 
Then  planes  ABC  and  BOB'  are  ±.  (§  441) 

Now  a  tangent  to  arc  AB  at  jB  is  J_  BOB'.  (§  439) 

Then  it  is  J_  to  a  tangent  to  arc  BB'  at  B.  (§  398) 

Then,  spherical  Z  ABB  is  a  rt.  Z.  (§  583) 

That  is,  an  arc  of  a  great  circle  drawn  from  the  pole  of  a 
great  circle  is  perpendicular  to  its  circumference. 

586.  Cor.  II.  The  angle  between  two  arcs  of  great  circles 
is  the  plane  angle  of  the  diedral  angle  between  their  planes. 

(§  429) 

SPHERICAL  POLYGONS  AND   SPHERICAL  PYRAMIDS. 
Definitions. 

587.  A  spherical  polygon  is  a  portion 
of  the  surface  of  a  sphere  bounded  by 
three  or  more  arcs  of  great  circles;  as 
ABCD. 

The  bounding  arcs  are  called  the  sides 
of  the  spherical  polygon,  and  are  usually 
measured  in  degrees. 

The  angles  of  the  spherical  polygon  are  the  spherical 
angles  (§  583)  between  the  adjacent  sides,  and  their  verti- 
ces are  called  the  vertices  of  the  spherical  polygon. 

A  diagonal  of  a  spherical  polygon  is  an  arc  of  a  great 
circle  joining  any  two  vertices  which  are  not  consecutive. 

A  spherical  triangle  is  a  spherical  polygon  of  three  sides. 

A  spherical  triangle  is  called  isosceles  when  it  has  two 
sides  equal ;  equilateral  when  all  its  sides  are  equal ;  and 
right-angled  when  it  has  a  right  angle. 

588.  The  planes  of  the  sides  of  a  spherical  polygon  form 
a  polyedral  angle,  whose  vertex  is  the  centre  of  the  sphere, 
and  whose  face  angles  are  measured  by  the  sides  of  the 
spherical  polygon  (§  192). 


I 


THE   SPHERE. 


337 


Thus,  in  the  figure  of  §  587,  the  planes  of  the  sides  of 
the  spherical  polygon  form  a  polyedral  angle,  0-ABCD, 
whose  face  AAOB,  BOC,  etc.,  are  measured  by  arcs  AB, 
BC,  etc.,  respectively. 

A  spherical  polygon  is  called  convex  when  the  polyedral 
angle  formed  by  the  planes  of  its  sides  is  convex  (§  453). 

589.  A  spherical  pyramid  is  a  solid  bounded  by  a  spherical 
polygon  and  the  planes  of  its  sides;  as  0-ABCD,  figure  of 
§  587. 

The  centre  of  the  sphere  is  called  the  vertex  of  the  spheri- 
cal pyramid,  and  the  spherical  polygon  the  base. 

Two  spherical  pyramids  are  equal  when   their  bases  are 


For  they  can  evidently  be  applied  one  to  the  other  so  as 
to  coincide  throughout. 

590.  If  circumferences  of  great  circles  be  drawn  with 
the  vertices  of  a  spherical  triangle  as  poles,  they  divide  the 
surface  of  the  sphere  into  eight  spherical  triangles. 

Thus,  if  circumference  B'C'B"  be 
drawn  with  vertex  A  of  spherical 
A  ABC  as  a  pole,  circumference 
A'C'A"  with  5  as  a  pole,  and  circum- 
ference A'B"A"B'  with  C  as  a  pole, 
the  surface  of  the  sphere  is  divided 
into  eight  spherical  A;  A'B'C, 
A'B"C',  A"B'C',  and  A"B"C'  on  the 
hemisphere  represented  in  the  figure,  the  others  on  the  oppo- 
site hemisphere. 

Of  these  eight  spherical  A,  one  is  called  the  polar  triangle 
of  ABC,  and  is  determined  as  follows : 

Of  the  intersections,  A'  and  A",  of  circumferences  drawn 
with  B  and  C  as  poles,  that  which  is  nearer  (§  573)  to  A, 
i.e.,  ^',  is  a  vertex  of  the  polar  triangle;  and  similarly  for 
the  other  intersections. 

Thus,  A'B'C  is  the  polar  A  of  ABC. 


338  SOLID   GEOMETRY.— BOOK  VIII. 

591.  Two  spherical  polygons,  on  the  same  or  equal 
spheres,  are  said  to  be  symmetrical  when  the  sides  and  an- 
gles of  one  are  equal,  respectively,  to  the  sides  and  angles 
of  the  other,  if  the  equal  parts  occur  in  the  reverse  order. 

Thus,  if  spherical  A  ABC  and 
A'B'C,  on  the  same  or  equal  spheres, 
have  sides  AB,  BG,  and  CA  equal,       A^C'  ^i 

respectively,  to  sides  A'B',  B'C,  and  B^  ^C 

CA',  and  A  A,  B,  and  C  to  A  A',  B',  and  C,  and  the  equal 
parts  occur  in  the  reverse  order,  the  A  are  symmetrical. 

It  is  evident  that,  in  general,  two  symmetrical  spherical 
polygons  cannot  be  placed  so  as  to  coincide  throughout. 

Prop.  XVI.     Theorem. 

592.  If  one  spherical  triangle  is  the  polar  triangle  of  an- 
other, then  the  second  spherical  triangle  is  the  polar  triangle 
of  the  first. 

A' 


Given  A'B'C  the  polar  A  of  spherical  A  ABC;  A,  B,  and 
C  being  the  poles  of  arcs  B'C,  CA',  and  A'B',  respectively. 

To  Prove  ABC  the  polar  A  of  spherical  A  A'B'C. 

Proof.     B  is  the  pole  of  arc  A'C. 

Whence,  A'  lies  at  a  quadrant's  distance  from  B.    (§  577) 

Again,  C  is  the  pole  of  arc  A'B'. 

Whence,  A'  lies  at  a  quadrant's  distance  from  C. 

Therefore,  A'  is  the  pole  of  arc  BC.  (§  578) 

Similarly,  B'  is  the  pole  of  arc  CA,  and  C  of  arc  AB. 

Then,  ABC  is  the  polar  A  of  A'B'C. 


THE   SPHERE.  33^ 

For  of  the  two  intersections  of  the  circumferences  having 
5'  and  C,  respectively,  as  poles,  A  is  the  nearer  to  A!)  and 
similarly  for  the  other  vertices.  (§  59O) 

Note.  Two  spherical  triangles,  each  of  which  is  the  polar  triangle 
of  the  other,  are  called  polar  triangles. 

Prop.  XVII.     Theorem. 

593.  In  two  polar  triangles,  each  angle  of  one  is  measured 
by  the  supplement  of  that  side  of  the  other  of  which  it  is  the 
pole. 

X 


JD 

Given  A,  B,  C,  A',  B',  and  C  the  A,  expressed  in  degrees, 
of  polar  A  ABO  and  A'B'C ;  A  being  the  pole  of  B'O, 
B  of  C'A',  C  of  A'B\  A'  of  BC,  B'  of  CA,  and  C  of  AB. 

Let  sides  BC,  CA,  AB,  B'C,  C'A',  and  A'B',  expressed 
in  degrees,  be  denoted  by  a,  b,  c,  a',  b',  and  c',  respectively. 
To  Prove 

A  =  180°  -  a',        B  =  180°  -b',        C  =  180°  -  c', 
A'  =  180°  -  a,         B'  =  180°  -b,         C"  =  180°  -  c. 
Proof.     Produce  arcs  J^  and  AC  to  meet  sltcB'C  at  D 
and  E,  respectively. 

Since  B'  is  the  pole  of  arc  AE,  and  C  of  arc  AD,  arcs  B'E 
and  CD  are  quadrants.  (§  577) 

.•.  arc  5'^  + arc  CD  =180°. 
Or,  arc  DE  +  arc  B'C  =  180°. 

But  since  A  is  the  pole  of  arc  B'C,  arc  DE  is  the  measure 
ofZA.  (§584) 

.'.  A  +  a'  =  180°,  or  ^  =  180°  -  a'. 
In  like  manner,  the  theorem  may  be  proved  for  any  Z  of 
either  A. 


340  SOLID   GEOMETRY.  — BOOK   VIII. 


Prop.  XVIII.     Theorem. 

594.   Any  side  of  a  spherical  triangle  is  less  than  the  sum 
of  the  other  two  sides. 


Given  AB  any  side  of  spherical  A  ABC. 

To  Prove  AB<AC  +  Ba 

(By  §  4.57,ZA0B<ZA0C-\-/:B0C',  and  these  Zs  are 
measured  by  sides  AB,  AC,  and  BC,  respectively.) 

Prop.  XIX.     Theorem. 

595.    The  sum  of  the  sides  of  a  convex  spherical  polygon  is 
less  than  360°. 

B 


Given  convex  spherical  polygon  ABCD. 
To  Prove       AB  +  BC+CD  +  DA<  360°. 

(By  §  458,  sum  of  AAOB,  BOC,  COD,  and  DOA  is 
<  360°.) 

Prop.  XX.     Theorem. 

596.    The  sum  of  the  angles  of  a  spherical  triangle  is  greater 
than  two,  and  less  than  six,  right  angles. 


341 


B' 

a' 

Given  A,  B,  and  C  the  A,  expressed  in  degrees,  of  spheri- 
cal A  ABC. 

To  Prove      A-{-B+C>  180°,  and  <  540°. 
Proof.     Let  A'B'O  be  the  polar  A  of  spherical  A  ABC, 
A  being  the  pole  of  B'O,  B  of  CA',  and  C  of  A^B'. 

Also,  let  sides  B'C,  C'A',  and  A'B',  expressed  in  degrees, 
be  denoted  by  a',  b',  and  c',  respectively. 
Then,  A  =  180°  -  a', 

B  =  180°  -  6', 

C  =  180°  -  c'.  (§  593) 

Adding  these  equations,  we  have 

A  +  J5+  C=  540°  -  (a'  +  6'  +  c').  (1) 

.-.  ^  +  5+C<540°. 
Again,  a'  +  6'  +  c'  <  360°.  (§  595) 

Whence,  by  (1),     A  +  B  +  C>  180°. 

597.  Cor.  A  spherical  triangle  may  have  one,  two,  or 
tlwee  right  angles,  or  one,  two,  or  three  obtuse  angles. 

DEFINITIONS. 

598.  A  spherical  triangle  having  two  right  angles  is 
called  2b  bi-rectangular  triangle,  and  one  having  three  right 
angles  a  tri-rectangular  triangle. 

599.  Two  spherical  polygons  on  the  same  sphere,  or 
equal  spheres,  are  said  to  be  mutually  equilateral,  or  mutu- 
ally equiangular,  when  the  sides  or  angles  of  one  are  equal, 
respectively,  to  the  homologous  sides  or  angles  of  the  other, 
whether  taken  in  the  same  or  in  the  reverse  order. 


342  SOLID   GEOMETRY.— BOOK  VIII. 

Prop.  XXI.     Theo:^em. 

600.  If  two  spherical  triangles  on  the  same  sphere,  or  equal 
spheres,  are  mutually  equiangular,  their  polar  triangles  are 
mutually  equilateral. 

A'  D' 


Given  ABC  and  DEF  mutually  equiangular  spherical  A 
on  the  same  sphere,  or  equal  spheres,  A  A  and  D  being 
homologous;  also,  A'B'C  the  polar  A  of  ABC,  and  D'E'F' 
of  DEF,  A  being  the  pole  of  B'C,  and  D  of  E'F'. 

To  Prove  A'B'C  and  D'E'F'  mutually  equilateral. 

Proof.  A  A  and  D  are  measured  by  the  supplements  of 
sides  B'C  and  E'F',  respectively.  (§593) 

But  by  hyp.,  ZA  =  AD. 

.-.  B'C  =  E'F.  (§  31,2) 

In  like  manner,  any  two  homologous  sides  of  A'B'C  and 
D'E'F'  may  be  proved  equal. 

Then,  A'B'C  and  D'E'F'  are  mutually  equilateral. 

601.  Cor.  (Converse  of  Prop.  XXI.)  If  two  spherical  tri- 
angles on  the  same  sphere,  or  equal  spheres,  are  mutually 
equilateral,  their  polar  triangles  are  mutually  equiangular. 

(The  proof  is  left  to  the  pupil ;  compare  §  600.) 

Prop.  XXII.     Theorem. 

602.  If  two  spherical  triangles  on  the  same  sphere,  or  equal 
spheres,  have  two  sides  and  the  included  angle  of  one  equal, 
respectively,  to  tioo  sides  and  the  included  angle  of  the  other, 

I.  They  are  equal  if  the  equal  parts  occur  in  the  same 
order. 


THE   SPHERE. 


343 


II.    They  are  symmetrical  if  the  equal  parts  occur  in  the 
reverse  order. 


i>' 


I.  Given  ABC  and  DEF  spherical  A  on  the  same  sphere, 
or  equal  spheres,  having 

AB  =  DE,  AC  =  DF,  and  ZA  =  ZD', 

the  equal  parts  occurring  in  the  same  order. 

To  Prove  A  ABC  =  A  DEF. 

Proof.  Superpose  A  ABC  upon  A  DEF  in  such  a  way 
that  Z  A  shall  coincide  with  its  equal  Z  D ;  side  AB  fall- 
ing on  side  DE,  and  side  AC  on  side  DF. 

Then,  since  AB  =  DE  and  AC  =  DF,  point  B  will  fall 
on  point  E,  and  point  C  on  point  F. 

Whence,  arc  BC  will  coincide  with  arc  EF.  ■  (§  572) 

Hence,  ABC  and  DEF  coincide  throughout,  and  are  equal. 

II.  Given  ABC  and  D'E'F'  spherical  A  on  the  same 
sphere,  or  equal  spheres,  having 

AB  =  D'E',  AC  =  D'F',  and  Z  ^  =  Z  Z)' ; 
the  equal  parts  occurring  in  the  reverse  order. 
To  Prove       ABC  and  D'E'F'  symmetrical. 
Proof.     Let  DEF  be  a  spherical  A  on  the  same  sphere, 
or  an  equal  sphere,  symmetrical  to  D'E'F,  having 
DE  =  D'E',  DF  =  D'F',  and  ZD  =  ZD'; 
the  equal  parts  occurring  in  the  reverse  order. 
Then,  in  spherical  A  ABC  and  DEF,  we  have 
AB  =  DE,  AC  =  DF,  and  ZA  =  Z.D', 
and  the  equal  parts  occur  in  the  same  order.  (Ax.  1) 

.  •.  A  ABC  =  A  DEF.  (§  602,  I) 

Therefore,  A  ABC  is  symmetrical  to  A  D'E'F'. 


344  SOLID   GEOMETRY.— BOOK  VIII. 


Prop.  XXIII.     Theorem. 

603.  If  two  spherical  triangles  on  the  same  sphere,  or  equal 
spheres,  have  a  side  and  two  adjacent  angles  of  one  equal, 
respectively,  to  a  side  and  two  adjacent  angles  of  the  other, 

I.    They  are  equal  if  the  equal  parts  occur  in  the  same  order. 
II.    They  are  symmetrical  if  the  equal  parts  occur  in  the 
reverse  order. 

(The  proof  is  left  to  the  pupil ;  compare  §  602.) 

Prop.  XXIV.     Theorem. 

604.  If  two  spherical  triangles  on  the  same  sphere,  or  equal 
spheres,  are  mutually  equilateral,  they  are  mutually  equiangular. 


Given  ABC  and  DEF  mutually  equilateral  spherical  A 
on  equal  spheres ;  sides  BC  and  EF  being  homologous. 

To  Prove  ABC  and  DEF  mutually  equiangular. 

Proof.  Let  0  and  0'  be  the  centres  of  the  respective 
spheres,  and  draw  lines  OA,  OB,  OC,  O'D,  O'E,  and  O'F. 

Now  the  triedral  A  0-ABC  and  O'-DEF  have  their  ho- 
mologous face  A  equal.  (§  192) 
.-.  diedral  Z  0A=  diedral  Z  O'D.  (§  459) 

But  the  Z  between  arcs  AB  and  AC  is  the  plane  Z  of 

diedral  Z  OA,  and  the  Z  between  arcs  DE  and  DF  is  the 

plane  Z  of  diedral  Z  O'D.  (§  586) 

.  ••  ZBAC  =  Z  EDF.  (§  434) 

In  like  manner,  any  two  homologous  A  of  ABC  and  DEF 
may  be  proved  equal. 

Whence,  ABC  and  DEF  are  mutually  equiangular. 


THE   SPHERE. 


345 


Note.  The  theorem  may  be  proved  in  a  similar  manner  when  the 
given  spherical  ^  are  on  the  same  sphere. 

605.  Cor.  If  two  spherical  triangles  on  the  same  sphere, 
or  equal  spheres,  are  mutually  equilateral, 

1.  They  are  equal  if  the  equal  parts  occur  in  the  same  order. 

2.  They  are  symmetrical  if  the  equal  parts  occur  in  the 
reverse  order. 

Prop.  XXV.     Theorem. 

606.  If  two  spherical  triangles  on  the  same  sphere,  or  equal 
spheres,  are  mutually  equiangular,  they  are  mutually  equi- 
lateral. 

A'  D' 


Given  ABC  and   DEF  mutually  equiangular  spherical 
A  on  the  same  sphere,  or  equal  spheres. 

To  Prove  ABC  and  DEF  mutually  equilateral. 

Proof.     Let  A'B'C  be  the  polar  A  of  ABC,  and  D'E'F 
of  DEF. 

Then,  since  ABC  and  DEF  are  mutually  equiangular, 
A'B'C  and  D'E'F'  are  mutually  equilateral.  (§  600) 

Then  A'B'C  and  D'E'F'  are  mutually  equiangular. 

(§  604) 

But  ABC  is  the  polar  A  of  A'B'C,  and  DEF  of  D'E'F'. 

(§  592) 

Then  ABC  and  DEF  are  mutually  equilateral.       (§  600) 

607.   Cor.  I.     If  two  spherical  triangles  on  the  same  sphere, 
or  equal  spheres,  are  mutually  equiangular, 

1.  They  are  equal  if  the  equal  parts  occur  in  the  same  order. 

2.  They  are  symmetrical  if  the  equal  parts  occur  in  the 
reverse  order. 


346 


SOLID   GEOMETRY.— BOOK  VIII. 


608.  Cor.  II.  If  three  diameters  of  a 
sphere  he  so  drawn  that  each  is  perpen- 
dicular to  the  other  two,  the  plaries  deter- 
mined by  them  divide  the  surface  of  the 
sphere  into  eight  equal  tri-rectangular  tri- 
angles. 

(Prove  by  §  607,  1.  By  §  585,  each 
Z  of  each  spherical  A  is  a  rt.  Z.) 

609.  Cor.  m.     The  surface  of  a  sphere  is  eight  times  the 
surface  of  one  of  its  tri-rectangular  triangles. 


Prop.  XXVI.    Theorem. 

610.   In  an  isosceles  spherical  triangle  the  angles  opposite 
the  equal  sides  are  equal. 


Given,  in  spherical  A  ABC,  AB  =  AC. 
To  Prove  ZB  =  ZC. 

Proof.     Draw  AD  an  arc  of  a  great  O,  bisecting  side  BC 
atZ). 

In  spherical  A  ABD  and  ACD,  AD  =  AD. 
Also,  AB  =  AC  and  BD  =  CD. 

Then,  ABD  and  ACD  are  mutually  equiangular.     (§  604) 
.-.  ZB  =  ZC. 

611.  Cor.  I.     An  isosceles  spherical  triangle  is  equal  to  the 
spherical  triangle  which  is  symmetrical  to  it. 

Por  the  equal  parts  occur  in  the  same  order. 

612.  Cor.  n.     (Converse  of  Prop.  XXVI.)     If  two  angles 
of  a  spherical  triangle  are  equal,  the  sides  opposite  are  equal. 


THE  SPHERE. 


347 


Given,  in  spherical  A  ABC,  Z.B  =  Z  C. 

To  Prove         AB  =  AC. 

Proof.  Let  A'B'C  be  the  polar  A  of 
ABC',  B  being  the  pole  of  A'O,  and  C  of 
A^B\  IB 

Then,  A'C  is  the  sup.  of  Z  B,  and  A^B'   ^ 

^^  ^  ^'  (§  593) 

.-.  A'C'  =  A'B'.  (§31,2) 

••.  ^B'  =  Z  C.  (§  610) 

But  ABC  is  the  polar  A  of  A'B'C;  B'  being  the  pole  of 
AC,  and  C  of  AB.  (§  592) 

Then  AB  is  the  sup.  of  Z  C,  and  ^0  of  Z  B'.  (?) 

.-.  ^^  =  ^(7.  .  (?) 


Prop.  XXVII.     Theorem. 

613.  If  two  angles  of  a  spherical  triangle  are  unequal,  the 
sides  opposite  are  unequal,  and  the  greater  side  lies  opposite 
the  greater  angle. 


Given,  in  spherical  A  ABC,  Z  ABC  >  Z  C. 
To  Prove  AC  >  AB. 

(Prove  by  a  method  analogous  to  that  of  §  99.  Draw  BD 
an  arc  of  a  great  O  meeting  AC  at  D,  and  making  Z  CBD 
equal  to  Z  C.) 

614.  Cor.  (Converse  of  Prop.  XXVII.)  If  two  sides  of  a 
spherical  triangle  are  unequal,  the  angles  opposite  are  unequal, 
and  the  greater  angle  lies  opposite  the  greater  side. 

(Prove  by  Beductio  ad  Absurdum.) 


348  SOLID   GEOMETRY. —BOOK  VIII. 


Prop.  XXVIII.     Theorem. 

615.  The  shortest  line  on  the  surface  of  a  sphere  between 
two  given  points  is  the  arc  of  a  great  circle,  not  greater  than  a 
semi-cii'cumference,  which  joins  the  points. 


Given  points  A  and  B  on  the  surface  of  a  sphere,  and  AB 
an  arc  of  a  great  O,  not  >  a  semi-circumference. 

To  Prove  AB  the  shortest  line  on  the  surface  of  the 
sphere  between  A  and  B. 

Proof.     Let  C  be  any  point  in  arc  AB. 

Let'  DCF  and  ECG  be  arcs  of  small  ©  with  A  and  B, 
respectively,  as  poles,  and  AC  and  BC  as  polar  distances. 

Now  arcs  DCF  and  ECG  have  only  point  C  in  common. 

For  let  F  be  any  other  point  in  arc  DCF,  and  draw  arcs 
of  great  CD  AF  and  BF. 

.'.  AF=Aa  (§  575) 

But,  AF+BF>AC-\-  BC.  (§  594) 

Subtracting  arc  AF  from  the  first  member  of  the  inequal- 
ity, and  its  equal  arc  AC  from  the  second  member, 

BF  >  BC,  or  BF  >  BG.  (§  575) 

Whence,  F  lies  without  small  O  ECG,  and  arcs  DCF  and 
ECG  have  only  point  C  in  common. 

We  will  next  prove  that  the  shortest  line  on  the  surface 
of  the  sphere  from  ^  to  jB  must  pass  through  C 

Let  ADEB  be  any  line  on  the  surface  of  the  sphere 
between  A  and  B,  not  passing  through  C,  and  cutting  arcs 
DCF  and  ECG  at  D  and  E,  respectively. 

Then,  whatever  the  nature  of  line  AD,  it  is  evident  that 
an  equal  line  can  be  drawn  from  A  to  C. 


THE   SPHERE.  349 

In  like  manner,  whatever  the  nature  of  line  BE,  an  equal 
line  can  be  drawn  from  B  to  C. 

Hence,  a  line  can  be  drawn  from  A  to  B  passing  through 
C,  equal  to  the  sum  of  lines  AD  and  BE,  and  consequently 
<  line  ADEB  by  the  portion  DE. 

Therefore,  no  line  which  does  not  pass  through  C  can  be 
the  shortest  line  between  A  and  B. 

But  by  hyp.,  C  is  any  point  in  arc  AB. 

Hence,  the  shortest  line  from  A  to  B  must  pass  through 
every  point  of  AB. 

Then,  the  arc  of  a  great  O  AB  is  the  shortest  line  on  the 
surface  of  the  sphere  between  A  and  B. 

EXERCISES. 

2.  If  the  sides  of  a  spherical  triangle  are  77°,  123°,  and  95°,  how 
many  degrees  are  there  in  each  angle  of  its  polar  triangle  ? 

3.  If  the  angles  of  a  spherical  triangle  are  86°,  131°,  and  68°,  how 
many  degrees  are  there  in  each  side  of  its  polar  triangle  ? 

MEASUREMENT  OF   SPHERICAL  POLYGONS. 
Definitions. 

616.  A  lune  is  a  portion  of  the  surface  of 
a  sphere  bounded  by  two  semi-circumfer- 
ences of  great  circles  ;  as  ACBD. 

The  angle  of  the  lune  is  the  angle  between 
its  bounding  arcs. 

617.  A  spherical  wedge  is  a  solid  bounded 
by  a  lune  and  the  planes  of  its  bounding  arcs. 

The  lune  is  called  the  base  of  the  spherical  wedge. 

618.  It  is  evident  that  two  lunes  on  the  same  sphere,  or 
equal  spheres,  are  equal  when  their  angles  are  equal 

619.  It  is  evident  that  two  spherical  ivedges  in  the  same 
sphere,  or  equal  spheres,  are  equal  when  the  angles  of  the  lanes 
which  form  their  bases  ai'e  equal. 


350  SOLID   GEOMETRY.  — BOOK   VIII. 

Prop.  XXIX.     Theorem. 

620.    The  sphencal  tnangles  corresponding   to   a  pair  of 
vertical  triedral  angles  are  symmetrical. 


Given  AOA',  BOB',  and  COC  diameters  of  sphere  AC; 
also,  the  planes  determined  by  them,  intersecting  the  sur- 
face in  circumferences  ABA'B',  BCB'C,  and  CACA'. 

To  Prove  spherical  A  ABC  and  A'B'C  symmetrical. 

Proof.  A  AOB,  BOC,  and  CO  A  are  equal,  respectively, 
to  AA'OB',  B'OC,  and  C'OA'.  (§  40) 

Then,  AB  =  A'B',  BC  =  B'C,  and  CA  =  CA'.         (§  192) 

But  the  equal  parts  of  ABC  and  A'B'C  occur  in  the 
reverse  order. 

Whence,  ABC  and  A'B'C  are  symmetrical.         (§  605,  2) 

Prop.  XXX.     Theorem. 

621.  Two  spherical  triangles  corresponding  to  a  pair  of 
vertical  triedral  angles  are  equivalent. 

A 


Given  AOA,  BOB',  and  COC  diameters  of  sphere  AB; 
also,  the  planes  determined  by  them,  intersecting  the  sur- 
face in  arcs  AB,  BC,  CA,  A'B',  B'C,  and  CA'. 


THE   SPHERE.  351 

To  Prove  area  ABC  =  area  A'B'C. 

Proof.  Let  P  be  the  pole  of  the  small  O  passing  through 
points  A,  B,  and  C,  and  draw  arcs  of  great  ®  PA,  PB, 
and  PC. 

.-.  PA  =  PB  =  PC.  (§575) 

Draw  the  diameter  of  the  sphere  PP',  and  the  arcs  of 
great  ®P'A',  P'B',  and  PC;  then,  spherical  A P^S  and 
P'A'B'  are  symmetrical.  (§  620) 

But  spherical  A  PAB  is  isosceles. 

.-.  A  PAB  =  A  P'A'B'.  (§  611) 

In  like  manner, 

A  PBC  =  A  PB'C  and  A  PCA  =  A  P'C'A'. 
Then  the  sum  of  the  areas  of  A  PAB,  PBC,  and  PCA 
equals  the  sum  of  the  areas  of  PA'B,  P'B'C,  and  PCA'. 
.'.  area  ABC  =  area  A'B'O. 

622.  Sch.  If  P  and  P'  fall  without  spherical  A  ABC  and 
A'B'C,  we  should  take  the  sum  of  the  areas  of  two  isos- 
celes spherical  A,  diminished  by  the  area  of  a  third. 

623.  Cor.  I.  Two  symmetrical  spherical  triangles  are  equiv- 
alent. 

624.  Cor.  II.  Spherical  pyramids  0-APB,  0-BPC,  and 
0-CPA  are  equal,  respectively,  to  spherical  pyramids 
0-A'PB',  O-B'PC,  and  0-CPA'.  (§  589) 

.-.  vol.  0-ABC=  vol.  0-A'B'C. 
Whence,  the  spherical  pyramids  corresponding  to  a  pair  of 
vertical  triedral  angles  are  equivalent. 

EXERCISES. 

4.  The  sura  of  the  angles  of  a  spherical  hexagon  is  greater  than  8, 
and  less  than  12,  right  angles.     (§  596.) 

5.  The  sum  of  the  angles  of  a  spherical  polygon  of  n  sides  is 
greater  than  2  w  —  4,  and  less  than  2  n^  right  angles. 

6.  The  arc  of  a  great  circle  drawn  from  the  vertex  of  an  isosceles 
spherical  triangle  to  the  middle  point  of  the  base,  is  perpendicular  to 
the  base,  and  bisects  the  vertical  angle. 


352 


SOLID   GEOMETRY.— BOOK  VIII. 


Prop.  XXXI.  Theorem. 

625.    Two  lunes  on  the  same  sphere,  or  equal  spheres,  are 
to  each  other  as  their  angles. 

Note.     The  word  ^^lune,'^  in  the  above  statement,  signifies  the 
area  of  the  lune. 

Case  I.    When  the  angles  are  commensurable. 


Given  ACBD  and  AGBE  lunes  on  sphere  AB,  having 
their  A  CAD  and  CAE  commensurable. 


To  Prove 


ACBD     Z.  CAD 


ACBE      Z  CAE 

Proof.  Let  /.  CAa  be  a  common  measure  of  A  CAD  and 
CAE,  and  let  it  be  contained  5  times  in  Z  CAD,  and  3  times 
in  Z  CAE. 

ACAD     5 


A  CAE     3 


(1) 


Producing  the  arcs  of  division  of  Z  CAD  to  B,  lune  ACBD 
will  be  divided  into  5  parts,  and  lune  ACBE  into  3  parts, 
all  of  which  parts  will  be  equal.  (§  618) 

ACBD     5 


Prom  (1)  and  (2), 


ACBE     3 

ACBD  ^  Z  CAD 
ACBE     A  CAE 


(2) 


(?) 


Note.    The  theorem  may  be  proved  in  a  similar  manner  when  the 
given  lunes  are  on  equal  spheres. 


THE   SPHERE.  353 

Case  II.    When  the  angles  are  incommenaurahle. 


(Prove  as  in  §§  189  or  244.  Let  Z  CAD  be  divided  into 
any  number  of  equal  parts,  and  apply  one  of  these  parts  to 
Z  CAE  as  a  unit  of  measure.) 

626.  Cor.  I.  The  surface  of  a  lune  is  to  the  surface  of  the 
sphere  as  the  angle  of  the  lune  is  to  four  right  angles. 

For  the  surface  of  a  sphere  may  be  regarded  as  a  lune 
whose  Z  is  equal  to  4  rt.  A. 

627.  Cor.  n.  If  the  unit  of  measure  for  angles  is  the  right 
angle,  the  area  of  a  lune  is  equal  to  twice  its  angle,  multiplied 
by  the  area  of  a  tri-rectangidar  triangle. 

Given  L  the  area  of  a  lune ;  A  the  numerical  measure  of 
its  Z  referred  to  a  rt.  Z  as  the  unit  of  measure ;  and  T  the 
area  of  a  tri-rectangular  A. 

To  Prove  L  =  2A  x  T. 

Proof.     The  area  of  the  surface  of  the  sphere  is  8  T. 

(§  609) 

.-.  -^  =  A  (§625) 

8r      4  ^         ^ 

.:  L  =  4x^T=2AxT. 
4 

628.  Sch.  I.  Let  it  be  required  to  find  the  area  of  a  lune 
whose  Z  is  50°,  on  a  sphere  the  area  of  whose  surface  is  72. 

The  Z  of  the  lune  referred  to  a  rt.  Z  as  the  unit  of 
measure  is  f ;  and  T  is  ^  of  72,  or  9. 

Then  the  area  of  the  lune  is  2  x  4  X  9,  or  10. 


354  SOLID  GEOMETRY.— BOOK   VIII. 

629.  Def .  A  tri-rectangular  pyramid  is  a  spherical  pyra- 
mid whose  base  is  a  tri-rectangular  triangle. 

630.  Sch.  n.     It  may  be  proved,  as  in  §  625,  that 

Two  spherical  tvedges  in  the  same  sphere,  or  equal  spheres, 
are  to  each  other  as  the  angles  of  the  lunes  which  form  their 
bases. 

(The  proof  is  left  to  the  pupil ;  see  §  619.) 

631.  Sch.  III.     It  may  be  proved  that 

If  the  unit  of  measure  for  angles  is  the  right  angle,  the 
volume  of  a  spherical  wedge  is  equal  to  twice  the  angle  of  the 
lune  which  forms  its  base,  multiplied  by  the  volume  of  a  tri- 
rectangular  pyramid. 

(The  proof  is  left  to  the  pupil ;  see  §§  626  and  627.) 

632.  Def.  The  sphencal  excess  of  a  spherical  triangle  is 
the  excess  of  the  sum  of  its  angles  above  180°  (§  596). 

Thus,  if  the  zi  of  a  spherical  A  are  65°,  80°,  and  95°,  its 
spherical  excess  is  65°  +  80°  +  95°  -  180°,  or  60°. 

Prop.  XXXII.     Theorem. 

633.  If  the  unit  of  measure  for  angles  is  the  right  angle, 
the  area  of  a  spherical  triangle  is  equal  to  its  spherical  excess, 
multiplied  by  the  area  of  a  tri-rectangular  triangle. 


Given  A,  B,  and  C  the  numerical  measures  of  the  A  of 
spherical  A  ABC,  referred  to  a  rt.  Z  as  the  unit  of  measure, 
and  T  the  area  of  a  tri-rectangular  A. 

To  Prove  area  ABC  =  (^  +  5  -f  O  -  2)  x  T. 


THE  SPHERE.  355 

Proof,     Complete  circumferences  ABA'B'f  ACA'C,  and 
BCB'C,  and  draw  diameters  AA\  BB',  and  CC. 
Then,  since  ABA'C  is  a  lune  whose  Z  is  A,  we  have 

area  ABC  +  area  ABC  =  2  ^  x  T  (§  626).  (1) 

And  since  BAB'C  is  a  lune  whose  Z  is  J5, 

area  ^5(7  +  area  AB'C  =2Bx  T.  (2) 

Again,  area  A'B'C  =  area  ^J5(7.  (§  620) 

Adding  area  ABC  to  both  members,  we  have 

area  ABC -\-  area  AB'C=  area  of  lune  CBCA 

=  2CxT.  (3) 

Adding  (1),  (2),  and  (3),  and  observing  that  the  sum  of 
the  areas  of  A  ABC,  A'BC,  AB'C,  and  A'B'C  is  equal  to 
the  area  of  the  surface  of  a  hemisphere,  or  4  T,  we  have 
2  area  ABC +  4:  T=  {2  A  +  2  B-\-2  C)  x  T. 
.-.  area  ABC-\-2  T={A  +  B+C)  x  T. 

.-.  area  ABC  =  (A-i- B  +  C -2)  x  T. 

634.  Sch.  I.  Let  it  be  required  to  find  the  area  of  a 
spherical  A  whose  A  are  105°,  80°,  and  95°,  on  a  sphere  the 
area  of  whose  surface  is  144. 

The  spherical  excess  of  the  spherical  A  is  100°,  or  J^  re- 
ferred to  a  rt.  Z  as  the  unit  of  measure ;  and  the  area  of  a 
tri-rectangular  A  is  |^  of  144,  or  18. 

Then  the  area  of  the  spherical  A  is  y-  x  18,  or  20. 

635.  Sch.  n.     It  may  be  proved,  as  in  §  633,  that 

If  the  unit  of  measure  for  angles  is  the  right  angle,  the 
volume  of  a  triangular  spherical  pyramid  is  equal  to  the 
spherical  excess  of  its  base,  multiplied  by  the  volume  of  a 
tri-rectangidar  pyramid. 

(The  proof  is  left  to  the  pupil ;  see  §§  624  and  630.) 

EXERCISES. 

7.  What  is  the  volume  of  a  spherical  wedge  the  angle  of  whose 
base  is  127°  30',  if  the  volume  of  the  sphere  is  112  ? 

8.  In  figure  of  Prop.  XVII.,  prove  A'  =  180°  -  a. 


356  SOLID   GEOMETRY, —BOOK   VIII. 

Prop.  XXXIII.     Theorem. 

636.  If  the  unit  of  measure  for  angles  is  the  right  angle, 
the  area  of  any  spherical  polygon  is  equal  to  the  sum  of  its 
angles,  diminished  by  as  many  times  two  right  angles  as  the 
figure  has  sides  less  two,  multiplied  by  the  area  of  a  tri- 
rectangular  triangle. 


Given  K  the  area  of  any  spherical  polygon,  n  the  number 
of  its  sides,  s  the  sum  of  its  A  referred  to  a  rt.  Z  as  the 
unit  of  measure,  and  T  the  area  of  a  tri-rectangular  A. 

To  Prove  K=  [s  -  2  (ii  -  2)]  x  T. 

Proof.  The  spherical  polygon  can  be  divided  into  ri  —  2 
spherical  A  by  drawing  diagonals  from  any  vertex. 

Now,  if  the  unit  of  measure  for  A  is  the  rt.  Z,  the  area 
of  each  spherical  A  is  equal  to  the  sum  of  its  A,  less  2  rt.  A, 
multiplied  by  T.  (§  633) 

Hence,  if  the  unit  of  measure  for  A  is  the  rt.  Z,  the  sum 
of  the  areas  of  the  spherical  A  is  equal  to  the  sum  of  their 
A,  diminished  by  n  —  2  times  2  rt.  A,  multiplied  by  T. 

But  the  sum  of  the  A  of  the  spherical  A  is  equal  to  the 
sum  of  the  A  of  the  spherical  polygon. 

Whence,  K=[s-2  (n  -  2)]  x  T. 

637.  Sch.  It  may  be  proved,  as  in  §  636,  that 
If  the  unit  of  measure  for  angles  is  the  right  angle,  the 
volume  of  any  spherical  pyramid  is  equal  to  the  sum  of  the 
angles  of  its  base,  diminished  by  as  many  times  two  right 
angles  as  the  base  has  sides  less  two,  multiplied  by  the  volume 
of  a  tri-recta7igular  pyramid. 
(The  proof  is  left  to  the  pupil.) 


THE   SPHERE.  357 


EXERCISES. 


9.   The  area  of  a  lune  is  28^.     If  the  area  of  the  surface  of  the 
sphere  is  120,  what  is  the  angle  of  the  lune  ? 

10.  Find  the  area  of  a  spherical  triangle  whose  angles  are  103°, 
112°,  and  127°,  on  a  sphere  the  area  of  whose  surface  is  160. 

11.  Find  the  volume  of  a  triangular  spherical  pyramid  the  angles 
of  whose  base  are  02°,  119°,  and  134°;  the  volume  of  the  sphere 
being  192. 

12.  What  is  the  ratio  of  the  areas  of  two  spherical  triangles  on  the 
same  sphere  whose  angles  are  94",  135°,  and  140°,  and  87°,  105°,  and 
118°,  respectively  ? 

13.  The  area  of  a  spherical  triangle,  two  of  whose  angles  are  78° 
and  99°,  is  34 1.  If  the  area  of  the  surface  of  the  sphere  is  234,  what 
is  the  other  angle  ? 

14.  The  volume  of  a  triangular  spherical  pyramid,  the  angles  of 
whose  base  are  105°,  126°,  and  147°,  is  60^  ;  what  is  the  volume  of  the 
sphere  ? 


15.    The  sides  opposite  the  equal  angles  of  a  bi- 
rectangular  triangle  are  quadrants.     (§  442.) 


16.  The  sides  of  a  spherical  triangle,  on  a  sphere 

the  area  of  whose  surface  is  156,  are  44°,  63°,  and  97°.     Find  the  area 
of  its  polar  triangle. 

17.  Find  the  area  of  a  spherical  hexagon  whose  angles  are  120°, 
139°,  148°,  155°,  162°,  and  167°,  on  a  sphere  the  area  of  whose  surface 
is  280. 

18.  Find  the  volume  of  a  pentagonal  spherical  pyramid  the  angles 
of  whose  base  are  109°,  128°,  137°,  153°,  and  158° ;  the  volume  of  the 
sphere  being  180. 

19.  The  volume  of  a  quadrangular  spherical  pyramid,  the  angles 
of  whose  base  are  110°,  122°,  135°,  and  146°,  is  12f ;  what  is  the 
volume  of  the  sphere  ? 

20.  The  area  of  a  spherical  pentagon,  four  of  whose  angles  are 
112°,  131°,  138°,  and  168°,  is  27.  If  the  area  of  the  surface  of  the 
sphere  is  120,  what  is  the  other  angle  ? 

21.  If  two  straight  lines  are  tangent  to  a  sphere  at  the  same  point, 
their  plane  is  t,angent  to  the  sphere.     (§  400.) 


358  SOLID   GEOMETRY. —BOOK  VIII. 

22.  The  sum  of  the  arcs  of  great  circles  drawn  from  any  point 
within  a  spherical  triangle  to  the  extremities  of  any  side,  is  less  than 
the  sum  of  the  other  two  sides  of  the  triangle. 

(Compare  §  48.) 

23.  How  many  degrees  are  there  in  the  polar  distance  of  a  circle, 
whose  plane  is  5\/2  units  from  the  centre  of  the  sphere,  the  diameter 
of  the  sphere  being  20  units  ? 

(The  radius  of  the  O  is  a  leg  of  a  rt.  A,  whose  hypotenuse  is  the 
radius  of  the  sphere,  and  whose  other  leg  is  the  distance  from  its 
centre  to  the  plane  of  the  O.) 

24.  The  chord  of  the  polar  distance  of  a  circle  of  a  sphere  is  6.  If 
the  radius  of  the  sphere  is  5,  what  is  the  radius  of  the  circle  ? 

D 

25.  If  side  AB  oi  spherical  triangle  ABC  is  a  / 
quadrant,  and  side  BC  less  than  a  quadrant,  prove 
Z^  less  than  90°. 


,^^, 


26.  The  polar  distance  of  a  circle  of  a  sphere  is  60°.  If  the 
diameter  of  the  circle  is  6,  find  the  diameter  of  the  sphere,  and  the 
distance  of  the  circle  from  its  centre. 

(Represent  radius  of  sphere  by  2  x.) 

27.  Any  point  in  the  arc  of  a  great  circle 
bisecting  a  spherical  angle  is  equally  distant 
(§  573)  from  the  sides  of  the  angle. 

(To  prove  arc  FM  =  arc  PiV.     Let  U  be 
a  pole  of  arc  AB,  and  F  of  arc  BC.     Spherical    ^"^ 
A  BFE  and  BFF  are  symmetrical  by  §  602, 
IL,  and  FE  =  FF.) 

28.  A  point  on  the  surface  of  a  sphere,  equally  distant  from  the 
sides  of  a  spherical  angle,  lies  in  the  arc  of  a  great  circle  bisecting 
the  angle. 

(Fig.  of  Ex.  27.  To  prove  Z  ABF  =  Z  CBF.  Spherical  A  BFE 
and  BFF  are  symmetrical  by  §  605,  2.) 

29.  The  arcs  of  great  circles  bisecting  the  angles  of  a  spherical 
triangle  meet  in  a  point  equally  distant  from  the  sides  of  the  triangle. 
(Exs.  27,  28,  p.  358.) 

30.  A  circle  may  be  inscribed  in  any  spherical  triangle. 

31.  State  and  prove  the  theorem  for  spherical  triangles  analogous 
to  Prop.  IX.,  L,  Book  I. 


THE   SPHEllE.  359 

32.  State  and  prove  the  theorem  for  spherical  triaugles  analogous 
to  Prop,  v.,  Book  I. 

33.  State  and  prove  the  theorem  for  spherical  triangles  analogous 
to  Prop.  L.,  Book  I.     (Ex.  32.) 

34.  If  PA,  PB,  and  PC  are  three  equal  arcs  of  great  circles  drawn 
from  point  P  to  the  circumference  of  great  circle  ABC,  prove  P  a  pole 
of  ABC. 

(PA  and  PB  are  quadrants  by  Ex.  15,  p.  357.) 

35.  The  spherical  polygons  corresponding  to  a  pair  of  vertical  poly- 
edral  angles  are  symmetrical.     (§  456.) 

36.  A  sphere  may  be  inscribed  in,  or  circumscribed  about,  any 
tetraedron.     (Ex.  73,  Book  VII.) 

37.  What  is  the  locus  of  points  in  space  at  a  given  distance  from 
a  given  straight  line  ? 

38.  Equal  small  circles  of  a  sphere  are  equally  distant  from  the 
centre. 

39.  State  and  prove  the  converse  of  Ex.  38. 

40.  The  less  of  two  small  circles  of  a  sphere  is  at  the  greater  dis- 
tance from  the  centre. 

41.  State  and  prove  the  converse  of  Ex,  40. 

42.  What  is  the  locus  of  points  on  the  surface  of  a  sphere  equally 
distant  from  the  sides  of  a  spherical  angle  ? 

43.  If  two  spheres  are  tangent  to  the  same  plane  at  the  same 
point,  the  straight  line  joining  their  centres  passes  through  the  point 
of  contact, 

44.  The  distance  between  the  centres  of  two  spheres  whose  radii 
are  25  and  17,  respectively,  is  28.  Find  the  diameter  of  their  circle 
of  intersection,  and  its  distance  from  the  centre  of  each  sphere. 

45.  If  a  polyedron  be  circumscribed  about  each  of  two  equal 
spheres,  the  volumes  of  the  polyedrons  are  to  each  other  as  the  areas 
of  their  surfaces. 

(Find  the  volume  of  each  polyedron  by  dividing  it  into  pyramids.) 

46.  Either  angle  of  a  spherical  triangle  is  greater  than  the  differ- 
ence between  180°  and  the  sum  of  the  other  two  angles. 

(Fig.  of  Prop.  XX,  To  prove  ZA>1S0°-(ZB  +  ZC),  or 
XZB  +  ZC)  -  180°,  according  asZ^+Z(7is<or>  180°.  In 
the  latter  case,  A'C  4-  A'B'>B'C';  then  use  §  593.) 


Book   IX. 


MEASUREMENT  OF  THE  CYLINDER,  CONE, 
AND  SPHERE. 

THE   CYLINDER. 
Definitions. 

638.  The  lateral  area  of  a  cylinder  is  the  area  of  its 
lateral  surface. 

A  right  section  of  a  cylinder  is  a  section  made  by  a  plane 
perpendicular  to  the  elements  of  its  lateral  surface. 

639.  A  prism  is  said  to  be  inscribed  in  a  cylinder  when  its 
lateral  edges  are  elements  of  the  cylindrical  surface. 

In  this  case,  the  bases  of  the  prism  are  inscribed  in  the 
bases  of  the  cylinder. 

A  prism  is  said  to  be  circumscribed  about  a  cylinder  when 
its  lateral  faces  are  tangent  to  the  cylinder,  and  its  bases  lie 
in  the  same  planes  with  the  bases  of  the  cylinder. 

In  this  case,  the  bases  of  the  prism  are  circumscribed 
about  the  bases  of  the  cylinder. 

640.  It  follows  from  §  363  that  ^^^ 

If  a  prism  whose  base  is  a  regular  KTT^     /~/V 

polygon   be   inscribed   in,  or   circum-  /  /p^zzzJ-^/ItJ 

scribed     about,    a    circular    cylinder  I  I  I    ffj/  I  / 

(§  540),  and  the  number  of  its  faces  11    /  '     I ! 

be  indefinitely  increased,  /  / //'..v/.-/::^^-./^/  / 

1.    The  lateral   area  of  the  prism  \['f    J     i/i^. 

approaches  the  lateral  area  of  the  cyl-  ^^■-/-^^ii>'*^^^'^^^ 

inder  as  a  limit.  n^-""^ 


MEASUREMENT  OF   THE   CYLINDER.  361 

2.  The  volume  of  the  prism  approaches  the  volume  of  the 
cylinder  as  a  limit. 

3.  The  perimeter  of  a  right  section  of  the  prism  approaches 
the  perimeter  of  a  right  section  of  the  cylinder  as  a  limit* 

Prop.  I.     Theorem. 

641.  The  lateral  area  of  a  circular  cylinder  is  equal  to  the 
perimeter  of  a  right  section  multiplied  by  an  element  of  the 
lateral  surface. 


Given  S  the  lateral  area,  P  the  perimeter  of  a  rt.  section, 
and  E  an  element  of  the  lateral  surface,  of  a  circular 
cylinder. 

To  Prove  S  =  P  x  E. 

Proof.  Inscribe  in  the  cylinder  a  prism  whose  base  is  a 
regular  polygon,  and  let  S'  denote  its  lateral  area,  and  P 
the  perimeter  of  a  rt.  section. 

Then,  since  the  lateral  edge  of  the  prism  is  E, 

S'  =  P'  X  E.  (§  484) 

Now  let  the  number  of  faces  of  the  prism  be  indefinitely 
increased. 

Then,  S'  approaches  the  limit  S, 

and  P  xE  approaches  the  limit  P  x  E.  (^  640, 1, 3) 

By  the  Theorem  of  Limits,  these  limits  are  equal.    (§  188) 
.-.  S  =  PxE. 

*  For  rigorous  proofs  of  these  statements,  see  Appendix,  p.  386. 


362  SOLID   GEOMETRY.  — BOOK  IX. 

642.  Cor.  I.  Tlie  lateral  area  of  a  cylinder  of  revolution 
is  equal  to  the  circumference  of  its  base  multiplied  by  its 
altitude. 

643.  Cor.  II.  If  S  denotes  the  lateral  area,  T  the  total 
area,  H  the  altitude,  and  R  the  radius  of  the  base,  of  a 
cylinder  of  revolution, 

S  =  27rEH.  (§368) 

And,     T=2 ttRH  +  2 ttR^  (§  371)  =  2 irR{H  +  R), 

Prop.  II.    Theorem. 

644.  The  volume  of  a  circular  cylinder  is  equal  to  the  prod- 
uct of  its  base  and  altitude. 


Given  V  the  volume,  B  the  area  of  the  base,  and  ^  the 
altitude,  of  a  circular  cylinder. 

To  Prove  V=BxH. 

Proof.  Inscribe  in  the  cylinder  a  prism  whose  base  is  a 
regular  polygon,  and  let  V  denote  its  Volume,  and  B'  the 
area  of  its  base. 

Then,  since  the  altitude  of  the  prism  is  H, 

r  =  B'xH.  (§499) 

Now  let  the  number  of  faces  of  the  prism  be  indefinitely 
increased. 

Then,  V  approaches  the  limit  V.  (§  640,  2) 

And,       B'  X  H  approaches  the  limit  B  x  H.     (§  363,  II) 
.-.   V=BxH.  (?) 


MEASUREMENT  OF  THE   CYLINDER. 


363 


645.  Cor.  If  V  denotes  the  volume,  H  the  altitude,  and 
li  the  radius  of  the  base,  of  a  circular  cylinder, 

V=  ttH^H.  (?) 

Prop.  III.     Theorem. 

646.  The  lateral  or  total  areas  of  two  similar  cylinders  of 
revolution  (§  550)  are  to  each  other  as  the  squares  of  their 
altitudes,  or  as  the  squares  of  the  radii  of  their  bases  ;  and 
their  volumes  are  to  each  other  as  the  cubes  of  their  altitudes^ 
or  as  the  cubes  of  the  radii  of  their  bases. 


Given  S  and  s  the  lateral  areas,  T  and  t  the  total  areas, 
V  and  V  the  volumes,  H  and  h  the  altitudes,  and  R  and  r 
the  radii  of  the  bases,  of  two  similar  cylinders  of  revolution. 

To  Prove    ^  =  ^=^'  =  ^,  and  ^=^  =  ^- 


Proof.     Since  the  generating  rectangles  are  similar, 

h       r 

H-\-R 


(§  253,  2) 


h  +  r 


(§  240) 


and     -^  = 


S^2jRH 

s        2  Trrh 
T^27rR{H+R) 
t         2Trr{h-\-r) 

V  ttR'H 

V  irT^h 


W 


(§643)=^x5  =  f  =  — 
r       r       r^       h^ 


(§  643) 


R^R 

r       r 


(§645)=^x^ 


R^ 


El 

'  h^' 


364  SOLID   GEOMETRY.— BOOK  IX. 

EXERCISES. 

1.  Find  the  lateral  area,  total  area,  and  volume  of  a  cylinder  of 
revolution,  the  diameter  of  whose  base  is  18,  and  whose  altitude  is  16. 

2.  The  radii  of  the  bases  of  two  similar  cylinders  of  revolution  are 
24  and  44,  respectively.  If  the  lateral  area  of  the  first  cylinder  is  720, 
what  is  the  lateral  area  of  the  second  ? 

3.  Find  the  altitude  and  diameter  of  the  base  of  a  cylinder  of 
revolution,  whose  lateral  area  is  168  tt  and  volume  504  tt. 

(Substitute  the  given  values  in  the  formulse  of  §§  643  and  646, 
and  solve  the  resulting  equations.) 

4.  Find  the  volume  of  a  cylinder  of  revolution,  whose  total  area 
is  170  IT  and  altitude  12. 

5.  How  many  cubic  feet  of  metal  are  there  in  a  hollow  cylindrical 
tube  18  ft.  long,  whose  outer  diameter  is  8  in.,  and  thickness  1  in.? 

(Find  the  difference  of  the  volumes  of  two  cylinders  of  revolution. 
TT  =3.1416.) 

6.  The  cross-section  of  a  tunnel,  2^  miles  in  length,  is  in  the  form 
of  a  rectangle  6  yd.  wide  and  4  yd.  high,  surmounted  by  a  semicircle 
whose  diameter  is  equal  to  the  width  of  the  rectangle  ;  how  many 
cu.  yd.  of  material  were  taken  out  in  its  construction  ?     (t  =  3.1416.) 

7.  The  volume  of  a  cylinder  of  revolution  is  equal  to  its  lateral 
area  multiplied  by  one-half  the  radius  of  its  base. 

THE  CONE. 
DEFINITIONS. 

647.  The  lateral  area  of  a  cone,  or  frustum  of  a  cone,  is 
the  area  of  its  lateral  surface. 

The  slant  height  of  a  cone  of  revolution  is  the  straight 
line  drawn  from  the  vertex  to  any  point  in  the  circumfer- 
ence of  the  base. 

The  slant  height  of  a  frustum  of  a  cone  of  revolution  is 
that  portion  of  the  slant  height  of  the  cone  included  between 
the  bases  of  the  frustum. 

648.  A  pyramid  is  said  to  be  inscribed  in  a  cone  when  its 
lateral  edges  are  elements  of  the  conical  surface ;  the  base 
of  the  pyramid  is  inscribed  in  the  base  of  the  cone,  and  its 
vertex  coincides  with  the  vertex  of  the  cone. 


MEASUREMENT  OF  THE   CONE.  3^5 

A  pyramid  is  said  to  be  circmnscribed  about  a  cone  when 
its  lateral  faces  are  tangent  to  the  cone,  and  its  base  lies  in 
the  same  plane  with  the  base  of  the  cone ;  the  base  of  the 
pyramid  is  circumscribed  about  the  base  of  the  cone,  and 
its  vertex  coincides  with  the  vertex  of  the  cone. 

649.  A  frustum  of  a  pyramid  is  said  to  be  inscribed  in  a 
fi'iistimi  of  a  cone  when  its  lateral  edges  are  elements  of  the 
lateral  surface  of  the  frustum  of  the  cone. 

In  this  case,  the  bases  of  the  frustum  of  the  pyramid  are 
inscribed  in  the  bases  of  the  frustum  of  the  cone. 

A  frustum  of  a  pyramid  is  said  to  be  circumscribed  about 
a  frustum  of  a  cone  when  its  lateral  faces  are  tangent  to  the 
frustum  of  the  cone,  and  its  bases  lie  in  the  same  planes 
with  the  bases  of  the  frustum  of  the  cone. 

In  this  case,  the  bases  of  the  frustum  of  the  pyramid  are 
circumscribed  about  the  bases  of  the  frustum  of  the  cone. 

650.  It  follows  from  §  363  that 

If  a  pyramid  whose  base  is  a  regidar  polygon  be  inscribed 
in,  or  circumscribed  about,  a  circular  cone 

(§  553),  and  the  number  of  its  faces  be  in-  /fnK 

definitely  increased,  /^ lv\ 

1.  The  lateral  area  of  the  pyramid  ap-  y/'y  M  I  •  •  \ 
proaches  the  lateral  area  of  the  cone  as  a  /^J-if^--h^^^^:\  \ 
limit.  X^/     /     I '  A> 

2.  The  volume  of  the  pyramid  approaches  n^^^^^^^sJ^^*^^ 
the  volume  of  the  cone  as  a  limit* 

651    It  follows  from  the  above  that 

If  a  frustum  of  a  pyramid  whose  base  is  a  regular  polygon 
be  inscribed  in,  or  circumscribed  about,  a  frustum  of  a  circu- 
lar cone,  and  the  number  of  its  faces  be  indefinitely  increased, 

1.  The  lateral  area  of  the  frustum  of  the  pyramid  ap- 
proaches the  lateral  area  of  the  frustum  of  the  cone  as  a  limit. 

2.  Tlie  volume  of  the  frustum  of  the  p?/r(xmid  approaches 
the  volume  of  the  frxistum  of  the  cone  as  a  limit. 

*  For  rigorous  proofs  of  these  statements,  see  Appendix,  p.  388. 


366  SOLID   GEOMETRY.  —  BOOK   IX. 


Prop.  IV.     Theorem. 

652.  The  lateral  area  of  a  cone  of  revolution  is  eqUal  to 
the  circumference  of  its  base,  multiplied  by  one-half  its  slant 
height. 


Given  S  the  lateral  area,  (7  the  circumference  of  the  base, 
and  L  the  slant  height,  of  a  cone  of  revolution. 

To  Prove  S  =  C  x  \  L. 

Proof.  Circumscribe  about  the  cone  a  regular  pyramid; 
let  aS'  denote  its  lateral  area,  and  C"  the  perimeter  of  its  base. 

Now  the  sides  of  the  base  of  the  pyramid  are  bisected  at 
their  points  of  contact  with  the  base  of  the  cone.       (§  174) 

Then,  the  slant  height  of  the  pyramid  is  the  same  as  the 

slant  height  of  the  cone.  (§  508) 

.-.  >S"=C"xiX.  (§  512) 

Now  let  the  number  of  faces  of  the  pyramid  be  indefi- 
nitely increased. 

Then,  /S"  approaches  the  limit  S.  (§  650,  1) 

And  C  x^L  approaches  the  limit  C  x  \  L.  (§  363,  I) 
.'.S=Cx\L.  (?) 

653.  Cor.  If  S  denotes  the  lateral  area,  T  the  total 
area,  L  the  slant  height,  and  R  the  radius  of  the  base,  of 
a  cone  of  revolution, 

S  =  2'!rItx\L{?)   =-kUL. 
And,  T  =  ttRL  -f  ttR''  (?)  =  7rR{L-\-  R). 

Prop.  V.     Theorem. 

654.  The  volume  of  a  circular  cone  is  equal  to  the  area  of 
its  base,  multiplied  by  one-third  its  altitude. 


MEASUREMENT  OF  THE   CONE.  367 


Given  V  tlie  volume,  B  the  area  of  the  base,  and  if  the 
altitude,  of  a  circular  cone. 
To  Prove  V=BxiH. 

(Inscribe  a  pyramid  whose  base  is  a  regular  polygon.) 

655.  Cor.  If  F  denotes  the  volume,  H  the  altitude,  and 
R  the  radius  of  the  base,  of  a  circular  cone, 

V=\.R'H.  (?) 

Prop.  VI.     Theorem. 

656.  Tlie  lateral  or  total  areas  of  tiuo  similar  cones  of  revo- 
lution are  to  each  other  as  the  squares  of  their  slant  heights,  or 
as  the  squares  of  their  altitudes,  or  as  the  squares  of  the  radii 
of  their  bases;  and  their  volumes  are  to  each  other  as  the  cubes 
of  their  slant  heights,  or  as  the  cubes  of  their  altitudes,  or  as 
cubes  of  the  radii  of  their  bases. 


Given  S  and  s  the  lateral  areas,  T  and  t  the  total  areas,  V 
and  V  the  volumes,  L  and  I  the  slant  heights,  H  and  h  the 
altitudes,  and  R  and  r  the  radii  of  the  bases,  of  two  similar 
cones  of  revolution  (§  555). 

To  Prove  ^=^=i^  =  ^'=^,  and  l=Il=El=^. 

s       t       I-       le       1^  V       P       ft"        r" 


368  SOLID   GEOMETRY.— BOOK  IX. 

(The  proof  is  left  to  the  pupil ;  compare  §  646.) 


Prop.  VII.     Theorem. 

657.  TJie  lateral  area  of  a  frusttim  of  a  cone  of  revolution 
is  equal  to  the  sum  of  the  circumferences  of  its  bases,  multiplied 
by  one-half  its  slant  height. 


Given  S  the  lateral  area,  C  and  c  the  circumferences  of 
the  bases,  and  L  the  slant  height,  of  a  frustum  of  a  cone  of 
revolution. 

To  Prove  S  =  (G  +  c)  x\L, 

Proof.  Circumscribe  about  the  frustum  of  the  cone  a 
frustum  of  a  regular  pyramid ;  let  >iS"  denote  its  lateral  area, 
and  O  and  c'  the  perimeters  of  its  bases. 

Now  the  sides  of  the  bases  of  the  frustum  of  the  pyra- 
mid are  bisected  at  their  points  of  contact  with  the  bases  of 
the  frustum  of  the  cone.  (§  174) 

Then,  the  slant  height  of  the  frustum  of  the  pyramid  is 
the  same  as  the  slant  height  of  the  frustum  of  the  cone. 

(§  508) 
.-.  ^'=((7'H-c')  x^L.  (§513) 

Now  let  the  number  of  faces  of  the  frustum  of  the 
pyramid  be  indefinitely  increased. 

Then,  /S"  approaches  the  limit  S,  (§  651,  1) 

and  (C  +  c')  X  ^L  approaches  the  limit  {C -\-c)  x\L. 

(§  363,  I) 
...  S={C  +  c)x\L.  (?) 


MEASUREMENT  OF   THE   CONE.  369 

658.  Cor.  I.  If  S  denotes  the  lateral  area,  L  the  slant 
height,  and  E  and  r  the  radii  of  the  bases,  of  a  frustum  of 
a  cone  of  revolution, 

S  =  (2TrE  +  2 Trr)  xiL(?)  =ir{R+  r)L. 

659.  Cor.  II.     We  may  write  the  first  result  of  §  668 

S  =  2'jrK\{R  +  r)xL. 

But,  2Tr  X  ^{R  +  r)  is  the  circumference  of  a  section 
equally  distant  from  the  bases.  (§  132) 

Whence,  the  lateral  area  of  a  frustum  of  a  coiie  of  revolu- 
tion is  equal  to  the  circumference  of  a  section  equally  distant 
from  its  bases,  multiplied  by  its  slant  height. 

Prop.  VIII.    Theorem. 

660.  Tlie  volume  of  a  frustum  of  a  circular  cone  is  equal 
to  the  sum  of  its  bases  and  a  mean  proportional  between  its 
bases,  multiplied  by  one-third  its  altitude. 


Given  V  the  volume,  B  and  b  the  areas  of  the  bases,  and 
H  the  altitude,  of  a  frustum  of  a  circular  cone. 

To  Prove       F=  (jB  -f  &  +  VBxb)  x\H. 

(Inscribe  a  frustum  of  a  pyramid  whose  base  is  a  regular 
polygon.     Then  apply  §  524.) 

661.   Cor.     If  V  denotes  the  volume,  H  the  altitude,  and 
R  and  r  the  radii  of  the  bases,  of  a  frustum  of  a  circular  cone, 
B  =  ■7rR\  b  =  Trr",  and  Vslib  =  V^i^^V  =  irRr.      (?) 
Then, 
F=  (ttR^  +  Tr?'^  +  TrRr)  X  \H  =  ^TriR' -{■  r"  +  Rr) H. 


370  SOLID   GEOMETRY.— BOOK   IX. 

EXERCISES. 

8.  Find  the  lateral  area,  total  area,  and  volume  of  a  cone  of  revo- 
lution, the  radius  of  whose  base  is  7,  and  whose  slant  height  is  25. 

9.  Find  the  lateral  area,  total  area,  and  volume  of  a  frustum  of 
a  cone  of  revolution,  the  diameters  of  whose  bases  are  16  and  6,  and 
whose  altitude  is  12. 

10.  The  slant  heights  of  two  similar  cones  of  revolution  are  9  and 
15,  respectively.  If  the  volume  of  the  second  cone  is  625,  what  is  the 
volume  of  the  first  ? 

11.  Find  the  volume  of  a  cone  of  revolution,  whose  slant  height  is 
29  and  lateral  area  580  tt. 

12.  Find  the  lateral  area  of  a  cone  of  revolution,  whose  volume  is 
320  TT  and  altitude  15. 

13.  The  altitude  of  a  cone  of  revolution  is  27,  and  the  radius  of  its 
base  is  16.  What  is  the  diameter  of  the  base  of  an  equivalent  cylinder 
of  revolution,  whose  altitude  is  16  ? 

14.  The  area  of  the  entire  surface  of  a  frustum  of  a  cone  of  revolu- 
tion is  306  TT,  and  the  radii  of  its  bases  are  11  and  5.  Find  its  lateral 
area  and  volume. 

15.  The  volume  of  a  frustum  of  a  cone  of  revolution  is  6020  tt,  its 
altitude  is  60,  and  the  radius  of  its  lower  base  is  15.  Find  the  radius 
of  its  upper  base  and  its  lateral  area. 

16.  Find  the  altitude  and  lateral  area  of  a  cone  of  revolution, 
whose  volume  is  800  tt,  and  whose  slant  height  is.  to  the  diameter  of 
its  base  as  13  to  10. 

17.  The  total  areas  of  two  similar  cylinders  of  revolution  are  32 
and  162,  respectively.  If  the  volume  of  the  second  cylinder  is  1458, 
what  is  the  volume  of  the  first  ? 

(Let  X  and  y  denote  the  altitudes  of  the  cylinders.) 

18.  The  volumes  of  two  similar  cones  of  revolution  are  343  and  512, 
respectively.  If  the  lateral  area  of  the  first  cone  is  196,  what  is  the 
lateral  area  of  the  second  ? 

19.  A  cubical  piece  of  lead,  the  area  of  whose  entire  surface  is 
384  sq.  in.,  is  melted  and  formed  into  a  cone  of  revolution,  the  radius 
of  whose  base  is  12  in.    Find  the  altitude  of  the  cone. 

20.  A  tapering  hollow  iron  column,  1  in.  thick,  is  24  ft.  long,  10  in. 
in  outside  diameter  at  one  end,  and  8  in.  in  diameter  at  the  other; 
how  many  cubic  inches  of  metal  were  used  in  its  construction  ? 

(Find  the  difference  of  the  volumes  of  the  frustums  of  two  cones 
of  revolution,     tt  =  3.1416.) 


MEASUREMENT  OF   THE   SPHERE. 


371 


21.  U  the  altitude  of  a  cone  of  revolution  is  three-fourths  the 
radius  of  its  base,  its  volume  is  equal  to  its  lateral  area  multiplied 
by  one-fifth  the  radius  of  its  base. 


THE   SPHERE. 
DEFINITIONS. 

662.  A  zone  is  a  portion  of  the  surface  of  a  sphere  in- 
cluded between  two  parallel  planes. 

Tlie  circumferences  of  the  circles  which  bound  the  zone 
are  called  the  bases,  and  the  perpendicular  distance  between 
their  planes  the  altitude. 

A  zone  of  one  base  is  a  zone  one  of  whose  bounding  planes 
is  tangent  to  the  sphere. 

A  spherical  segment  is  a  portion  of  a  sphere  included  be- 
tween two  parallel  planes. 

The  circles  which  bound  it  are  called  the  bases,  and  the 
perpendicular  distance  between  them  the  altitude. 

A  spherical  segment  of  one  base  is  a  spherical  segment  one 
of  whose  bounding  planes  is  tangent  to  the  sphere. 

663.  If  semicircle  ACEB  be  revolved 
about  diameter  AB  as  an  axis,  and  CD 
and  EF  are  lines  ±  AB,  arc  CE  generates 
a  zone  whose  altitude  is  DF,  figure  CEFD 
a  spherical  segment  whose  altitude  is  DF, 
arc  AC  a  zone  of  one  base,  and  figure  ACD 
a  spherical  segment  of  one  base. 

664.  If  a  semicircle  be  revolved  about  its  diameter  as  an 
axis,  the  solid  generated  by  any  sector  of 
the  semicircle  is  called  a  spherical  sector. 

Thus,  if  semicircle  ACDB  be  revolved 
about  diameter  AB  as  an  axis,  sector  OCD 
generates  a  spherical  sector. 

The  zone  generated  by  the  arc  of  the 
sector  is  called  the  base  of  the  spherical 
sector. 


372 


SOLID  GEOMETRY  — BOOK  IX. 


Prop.  IX.     Theorem. 

665.  Tlie  area  of  the  surface  generated  by  the  revolution  of 
a  straight  line  about  a  straight  line  in  its  plane,  not  parallel  to 
and  not  intersecting  it,  as  an  axis,  is  equal  to  its  projection  on 
the  axis,  multiplied  by  the  circumference  of  a  circle,  whose 
radius  is  the  perpendicular  erected  at  the  middle  point  of  the 
line  and  terminating  in  the  axis. 


Given  str.  line  AB  revolved  about  str.  line  FM  in  its 
plane,  not  II  to'  and  not  intersecting  it,  as  an  axis;  lines 
AC  and  BD±FM,  and  EF  the  J_  erected  at  the  middle 
point  of  AB  terminating  in  FM. 

To  Prove  area  AB*  =  CD x  2 tt EF.     (§ §  276,  368) 

Proof.     Draw  line  AG  J_  BD,  and  line  EH  ±  CD. 
The  surface  generated  by  AB  is  the  lateral  surface  of  a 
frustum  of  a  cone  of  revolution,  whose  bases  are  generated 
by  AC  and  BD. 

.'.  area  AB  =  ABx2Tr  EH. 
But  A  ABG  and  EFH  are  similar. 
AB^EF 
AG      eh' 
.'.  AB  X  EH=  AG  X  EF 

=  CD  X  EF. 

Substituting,  we  have 

area  AB==CD  x2  7r  EF. 


(§  659) 
(§  262) 

(?) 

(§  232) 

(?) 


*  The  expression  " area  AB''''  is  used  to  denote  the  area  of  the  $i(r- 
face  generated  by  AB. 


MEASUREMENT  OF  THE  SPHERE.  373 

Prop.  X.     Theorem. 

666.  If  an  isosceles  triangle  he  revolved  about  a  straight 
line  in  its  plane,  not  parallel  to  its  base,  as  an  axis,  which 
passes  through  its  veHex  ivithout  intersecting  its  surface,  the 
volume  of  the  solid  generated  is  equal  to  the  area  of  the  surface 
generated  by  the  base,  multiplied  by  one-third  the  altitude. 


^0 
Given  isosceles  A  OAB  revolved  about  str.  line  OF  in  its 
plane,  not  II  to  base  AB,  as  an  axis  ;  and  line  OC 1.  AB. 
To  Prove     vol.  OAB  *  =  area  AB  x  \  OC. 
Proof.     Draw  lines  AD  and  BE  1.  OF;  and  produce  BA 
to  meet  OF  a^t  F. 

Now,  vol.  OBF  =  vol.  OBE  +  vol.  BEF 

=  \^BE' xOE  +  \itBE^xEF  {^Qb^) 

=  \  ttBE"  X  {OE  +  EF)  =  i  ttBE  X  be  X  OF. 

But  BE  X  OF  =  OC  X  BF,  for  each  expresses  twice  the 

area  of  AOBF.  (?) 

.-.  vol.  OBF  =  1  ttBE  xOCx  BF. 

But  ttBE  X  BF  is  the  area  of  the  surface  generated  by  BF. 

(§  653) 
.-.  vol.  OBF=  area  BF  x  ^  OC.  (1) 

Similarly,         vol.  OAF  =  area  AF  x  ^  OC.  (2) 

Subtracting  (2)  from  (1),  we  have 

vol.  OAB  =  (area  BF-  area  AF)  x\OC 
=  area  AB  x  \  OC. 

*  The  expression  "  vol,  OAB  "  is  used  to  denote  the  volume  of  the 
solid  generated  by  OAB. 


374  SOLID   GEOMETRY.— BOOK   IX. 

Prop.  XI.     Thp:orem. 

667.    The  area  of  a  zone  is  equal  to  its  altitude  multiplied 
by  the  circumference  of  a  great  circle. 

M 


A' 


C, 


Given  arc  AB  revolved  about  diameter  OM  as  an  axis, 
lines  AA^  and  BB'  J_  OM,  and  R  the  radius  of  the  arc. 
To  Prove  area  of  zone  generated  by  AB  =  A'B'  x  2  ttR. 
Proof.     Divide  arc  AB  into  three  equal  arcs,  AC,  CD, 
and  DB,  and  draw  chords  AC,  CD,  and  DB. 

Also,  draw  lines  CC  and  DD'±OM,  and  line  OE  ±  AC. 
.'.  area^(7  =  .4'(7'  x^irOE, 

area  CD  =  CD'  x  2  irOE,  etc.  (§  665) 

Adding  these  equations,  we  have 
area  of  surface  generated  by  broken  line  ACDB 

=  (^'C  +  C'Z)'  +  etc.)  X  2  ttOE  =  A'B'  x  2  ttOE. 
Now  let  the  subdivisions  of  arc  AB  be  bisected  indefinitely. 
Then,  area  of  surface  generated  by  broken  line  ACDB 
approaches  area  of  surface  generated  by  arc  AB  as  a  limit. 

(§363,1*) 
And,  A'B'  X  2  irOE  approaches  A'B'  x  2  ttJ?  as  a  limit. 

(§  364,  1*) 

*  The  broken  line  ACDB  is  called  a  regular  broken  line,  and  is  said 

to  be  inscribed  in  arc  AB ;  the  theorems  of  §§  363,  I,  and  364,  1,  are 

evidently  true  when,  instead  of  the  perimeter  of  a  regular  inscribed 

polygon,  we  have  a  regular  broken  line  inscribed  in  an  arc. 

For  a  rigorous  proof  of  the  statement  that  the  area  of  the  surface 
generated  by  ACDB  approaches  the  area  of  the  surface  generated  by 
arc  AB  as  a  limit,  see  Appendix,  p.  390. 


MEASUREMENT  OF  THE   SPHERE.  375 

Then,  area  of  zone  generated  by  arc  AB  =  A'B'  x  2  ttR. 

(§  188) 

668.  Sch.  The  proof  of  §  667  holds  for  any  zone  which 
lies  entirely  on  the  surface  of  a  hemisphere ;  for,  in  that 
case,  no  chord  is  II  OM,  and  §  665  is  applicable. 

Since  a  zone  which  does  not  lie  entirely  on  the  surface  of 
a  hemisphere  may  be  considered  as  the  sum  of  two  zones, 
each  of  which  does  lie  entirely  on  the  surface  of  a  hemi- 
sphere, the  theorem  of  §  667  is  true  for  any  zone. 

669.  Cor.  I.  If  S  denotes  the  area  of  a  zone,  h  its  alti- 
tude, and  E  the  radius  of  the  sphere, 

8  =  2^^1. 

670.  Cor.  II.  Since  the  surface  of  a  sphere  may  be  re- 
garded as  a  zone  whose  altitude  is  a  diameter  of  the  sphere, 
it  follows  that 

The  area  of  the  surface  of  a  sphere  is  equal  to  its  diameter 
multiplied  by  the  circumference  of  a  great  circle. 

671.  Cor.  m.  Let  S  denote  the  area  of  the  surface  of 
a  sphere,  M  its  radius,  and  D  its  diameter. 

Then,  S  =  2  R  x2  7rR(?)  =  A  irR'. 

That  is,  the  area  of  the  surface  of  a  sphere  is  equal  to  the 
square  of  its  radius  multiplied  by  4  tt. 

Again,  S  =  7rx(2Ry  =  ttI^. 

That  is,  the  area  of  the  surface  of  a  sphere  is  equal  to  the 
square  of  its  diameter  multiplied  by  tt. 

672.  Cor.  IV.  The  surface  of  a  sphere  is  equivalent  to 
four  great  circles. 

For  ttR^  is  the  area  of  a  great  O.  (?) 

673.  Cor.  V.  The  areas  of  the  su^r faces  of  two  spheres 
are  to  each  other  as  the  squares  of  their  radii,  or  as  the  squares 
of  their  diameters. 

(The  proof  is  left  to  the  pupil ;  compare  §  372.) 


376 


SOLID   GEOMETRY.— BOOK  IX. 


EXERCISES. 

22.  Find  the  area  of  the  surface  of  a  sphere  whose  radius  is  12. 

23.  Find  the  area  of  a  zone  whose  altitude  is  13,  if  the  radius  of 
the  sphere  is  16. 

24.  Find  the  area  of  a  spherical  triangle  whose  angles  are  125°, 
133°,  and  156°,  on  a  sphere  whose  radius  is  10. 

Prop.  XII.     Theorem. 

674.  The  volume  of  a  spherical  sector  is  equal  to  the  area 
of  the  zone  which  forms  its  base,  multiplied  by  one-third  the 
radius  of  the  sphere. 


Given  sector  OAB  revolved  about  diameter  OM  as  an 
axis,  and  R  the  radius  of  the  arc. 

To  Prove  volume  of  spherical  sector  generated  by  OAB 
=  area  of  zone  generated  by  AB  x\R. 

Proof.     Divide  arc  AB  into  three  equal  arcs,  AC,  CD,  and 
DB,  and  draw  chords  AC,  CD,  and  DB. 

Also,  draw  lines  OC  and  OD,  and  line  OE  Jl  AC. 
.-.  vol.  0^C=  area  AG  x  \  OE, 

vol.  OCD  =  area  CD  x  ^  OE,  etc.         (§  666) 
Adding  these  equations,  we  have 

volume  of  solid  generated  by  polygon  OACDB 
=  (area  AC  +  area  CD  +  etc.)  x  ^  OE 
=  area  ACDB  x  |  OE. 
Now  let  the  subdivisions  of  arc  AB  be  bisected  indefi- 
nitely. 


MEASUREMENT  OF  THE   SPHERE.  377 

Then,  volume  of  solid  generated  by  polygon  OACDB 
approaches  volume  of  solid  generated  by  sector  OAB  as  a 
limit.  (§  363,  II  *) 

And  area  of  surface  generated  by  ACDB  x  \  OE  ap- 
proaches area  of  surface  generated  by  arc  AB  x  ^  i2  as  a 
limit.  (§§  363,  I,  364,  1 1) 

Then,  volume  of  solid  generated  by  sector  OAB 

=  area  of  zone  generated  by  arc  AB  x  ^  R.         (?) 

675.  Sch.  It  is  evident,  as  in  §  668,  that  the  theorem  of 
§  674  holds  for  any  spherical  sector. 

676.  Cor.  I.  If  V  denotes  the  volume  of  a  spherical  sec- 
tor, h  the  altitude  of  the  zone  which  forms  its  base,  and  li 
the  radius  of  the  sphere, 

V=2'7rEh  X  ii2(§  669)  =  i^I^k. 

677.  Cor.  II.  Since  a  sphere  may  be  regarded  as  a 
spherical  sector  whose  base  is  the  surface  of  the  sphere, 

The  volume  of  a  sphere  is  equal  to  the  area  of  its  surface 
multiplied  by  one-third  its  radius. 

678.  Cor.  m.     Let  V  denote  the  volume  of  a  sphere,  E 

its  radius,  and  D  its  diameter.  • 

Then,  F  =  4  TT  i?2  X  i  i?  (§  671)  =  1 7ri2». 

That  is,  the  volume  of  a  sphere  is  equal  to  the  cube  of  its 
radius  multiplied  by  |  tt. 

Again,  F  =  7rZ>2  X  i  Z>  (§  671)  =  \  -kB^. 

That  is,  tlie  volume  of  a  sphere  is  equal  to  the  cube  of  its 
diameter  multiplied  by  \  tt. 

*  The  polygon  OACDB  is  called  a  regular  polygonal  sector^  and  is 
said  to  be  inscribed  in  sector  OAB  ;  the  theorem  of  §  363,  II,  is  evi- 
dently true  when,  instead  of  a  regular  inscribed  polygon,  we  have  a 
regular  polygonal  sector  inscribed  in  a  sector. 

For  a  rigorous  proof  of  the  statement  that  the  volume  of  the  solid 
generated  by  OA  CDB  approaches  the  volume  of  the  solid  generated 
by  sector  OAB  as  a  limit,  see  Appendix,  p.  391. 

t  See  note  foot  of  p.  374. 


378 


SOLID   GEOMETRY.— BOOK  IX. 


679.  Cor.  rV.  The  volumes  of  two  spheres  are  to  each 
other  as  the  cubes  of  their  radii,  or  as  the  cubes  of  their  diame- 
ters. 

(The  proof  is  left  to  the  pupil.) 

680.  Cor.  V.  The  volume  of  a  spherical  pyramid  is  equal 
to  the  area  of  its  base  multiplied  by  one-third  the  radius  of  the 
sphere. 

Given  P  the  volume  of  a  spherical  pyramid,  K  the  area 
of  its  base,  and  R  the  radius  of  the  sphere. 


To  Prove 


F=KxlB. 


Proof.  Let  n  denote  the  number  of  sides  of  the  base  of 
the  spherical  pyramid,  s  the  sum  of  its  A  referred  to  a  rt.  Z 
as  the  unit  of  measure,  T  the  area  of  a  tri-rectangular  A, 
T'  the  volume  of  a  tri-rectangular  pyramid,  S  the  area  of 
the  surface  of  the  sphere,  and  V  its  volume. 

P  _[s-2(n-2)-]  xT'  _T' 
ir~[s-2(7i-2)]  xT~  T' 

V  ^sr  ^r 

S      ST      t' 


Then, 


Also, 


(§§636,637) 
(§  609) 


P 
K 


V 

S' 


i- 


R^ 


4.TrR 


-,(§§671,  678)  =  ii^. 


P=K  x^R. 


Prop.  XIII.     Problem. 

681.    Given  the  radii  of  the  bases,  and  the  altitude,  of  a 
spherical  segment,  to  find  its  volume. 


MEASUREMENT  OF  THE   SPHERE.  379 

Given  O  the  centre  of  arc  ADB,  lines  AA  and  BB'  1.  to 
diameter  OM,  AA'  —  r\  BB'  =  r,  A'B'  =  h,  and  figure 
ADBB'A'  revolved  about  OM  as  an  axis. 

Required  to  express  volume  of  spherical  segment  gener- 
ated by  ADBB'A'  in  terms  of  r,  r',  and  A. 

Solution.  Draw  lines  OA,  OB,  and  AB ;  also,  line  OC  ± 
AB,  and  line  AE 1.  BB' ;  and  denote  radius  OA  by  R. 

Now,  vol.  ADBB'A'  =  vol.  ACBD  +  vol.  ABB' A'.       (1) 

Also,  •      vol.  ACBD  =  vol.  OADB  -  vol.  OAB. 

But,         vol.  OADB  =  I  TrR'h.  (§  676) 

And,  vol.  OAB  =  area  AB  x  i  OC  (§  666) 

=  h  x27rOCxiOC  (§665) 

.'.   vol.  ^Ci>i?  =  |7ri?%  -  I  TT  OO'A 
=  |7r(i22-  00")/^. 
But,  E'-OC'  =  AC'  (§  273) 

=  (i^y  (?) 

.-.  vol.  ACDB  =  I TT  X  i  Zb"  X  /i  =  i  ttZb"  Jl 
Now,  ^5"  =  ^:^'  +  AE"  (?) 

=  (,.  _  r'y  +  ^2  (?) 

.-.  vol.  ACDB  =  i  TT  [(r  -  r^  +  ^i^j;^^ 
Also,      vol.  ABB' A'  =  i  tt  (r^  +  r'^  +  rr')  /i.  (§  661) 

Substituting  in  (1),  we  have 
Yol.  ADBB'A' 

=  i  TT  [(?•  -  r'f  +  7^2]  /i  +  1 TT  (2r2  +  2r'2  +  2rr')  /i 


=  i- 


2rr'  +  r'^  +  7^2  _^  2  r^  +  2r'2  +  2rr')  h 


380  SOLID   GEOMETRY. —BOOK  IX. 

682.   Cor.     If  r  denotes  the  radius  of  the  base,  and  h  the 
altitude,  of  a  spherical  segment  of  one  base,  its  volume  is 
iTrr'h  +  l'rrh^ 

EXERCISES. 

25.  Find  the  volume  of  a  sphere  whose  radius  is  12. 

26.  Find  the  volume  of  a  spherical  sector,  the  altitude  of  whose 
base  is  12,  the  diameter  of  the  sphere  being  25. 

27.  Find  the  volume  of  a  spherical  segment,  the  radii  of  whose 
bases  are  4  and  5,  and  whose  altitude  is  9. 

28.  Find  the  radius  and  volume  of  a  sphere,  the  area  of  whose 
surface  is  324  w. 

29.  Find  the  diameter  and  area  of  the  surface  of  a  sphere  whose 
volume  is  M^  t, 

30.  The  surface  of  a  sphere  is  equivalent  to  the  lateral  surface  of 
its  circumscribed  cylinder. 

31.  The  volume  of  a  sphere  is  two-thirds  the  volume  of  its  circum- 
scribed cylinder. 

32.  A  spherical  cannon-ball  9  in.  in  diameter  is  dropped  into  a 
cubical  box  filled  with  water,  whose  depth  is  9  in.  How  many  cubic 
inches  of  water  will  be  left  in  the  box  ?     (ir  =  3.1416.) 

33.  What  is  the  angle  of  the  base  of  a  spherical  wedge  whose 
volume  is  ^^  tt,  if  the  radius  of  the  sphere  is  4  ? 

34.  Find  the  volume  of  a  quadrangular  spherical  pyramid,  the 
angles  of  whose  base  are  107°,  118°,  134°,  and  146° ;  the  diameter  of 
the  sphere  being  12. 

35.  The  surface  of  a  sphere  is  equivalent  to  two-thirds  the  entire 
surface  of  its  circumscribed  cylinder. 

36.  Prove  Prop.  IX.  when  the  straight  line  is  parallel  to  the  axis. 

37.  Find  the  area  of  the  surface  and -the  volume  of  a  sphere 
inscribed  in  a  cube  the  area  of  whose  surface  is  486. 

38.  How  many  spherical  bullets,  each  f  in.  in  diameter,  can  be 
formed  from  five  pieces  of  lead,  each  in  the  form  of  a  cone  of  revolu- 
tion, the  radius  of  whose  base  is  5  in.,  and  whose  altitude  is  8  in.  ? 

39.  A  cylindrical  vessel,  8  in.  in  diameter,  is  filled  to  the  brim  with 
water.  A  ball  is  immersed  in  it,  displacing  water  to  the  depth  of  2\  in. 
Find  the  diameter  of  the  ball. 


MEASUREMENT  OF  THE  SPHERE.        381 

40.  If  a  sphere  6  in,  in  diameter  weighs  351  ounces,  what  is  the 
weight  of  a  sphere  of  the  same  material  whose  diameter  is  10  in.  ? 

41.  If  a  sphere  whose  radius  is  12^  in.  weighs  3125  lb.,  what  is  the 
radius  of  a  sphere  of  the  same  material  whose  weight  is  819^  lb.  ? 

42.  The  altitude  of  a  frustum  of  a  cone  of  revolution  is  3J,  and  the 
radii  of  its  bases  are  5  and  3  ;  what  is  the  diameter  of  an  equivalent 
sphere  ? 

43.  Find  the  radius  of  a  sphere  whose  surface  is  equivalent  to  the 
entire  surface  of  a  cylinder  of  revolution,  whose  altitude  is  10^,  and 
radius  of  base  3. 

44.  The  volume  of  a  cylinder  of  revolution  is  equal  to  the  area  of 
its  generating  rectangle,  multiplied  by  the  circumference  of  a  circle 
whose  radius  is  the  distance  to  the  axis  from  the  centre  of  the 
rectangle. 

45.  The  volume  of  a  cone  of  revolution  is  equal  to  its  lateral  area, 
multiplied  by  one-third  the  perpendicular  from  the  vertex  of  the  right 
angle  to  the  hypoteniise  of  the  generating  triangle. 

46.  Two  zones  on  the  same  sphere,  or  equal  spheres,  are  to  each 
other  as  their  altitudes. 

47.  The  area  of  a  zone  of  one  base  is  equal  to  the  area  of  the  circle 
whose  radius  is  the  chord  of  its  generating  arc.     (§  270,  2.) 

48.  If  the  radius  of  a  sphere  is  B,  what  is  the  area  of  a  zone  of  one 
base,  whose  generating  arc  is  45°  ?     (Ex.  65,  p.  210.) 

49.  If  the  altitude  of  a  cone  of  revolution  is  15,  and 
its  slant  height  17,  find  the  total  area  of  an  inscribed 
cylinder,  the  radius  of  whose  base  is  5. 

(Let  the  cone  and  cylinder  be  generated  by  the  revo- 
lution of  rt.  A  ABC  and  rect.  CDEF  about  ^C  as  an 
axis.) 

50.  Find  the  area  of  the  surface  and  the  volume  of  a  sphere  cir- 
cumscribing a  cylinder  of  revolution,  the  radius  of  whose  base  is  9, 
and  whose  altitude  is  24. 

51.  An  equilateral  triangle,  whose  side  is  6,  revolves  about  one  of 
its  sides  as  an  axis.  Find  the  area  of  the  entire  surface,  and  the 
volume,  of  the  solid  generated. 

52.  A  cone  of  revolution  is  inscribed  in  a  sphere  whose  diameter 
is  I  the  altitude  of  the  cone.  Prove  that  its  lateral  surface  and  vol- 
ume are,  respectively,  |  and  /^  the  surface  and  volume  of  the  sphere. 


382 


SOLID   GEOMETRY. —BOOK  IX. 


53.    Find  the  volume  of  a  sphere  circumscribing  a  cube  whose 
volume  is  64. 


54.  A  cone  of  revolution  is  circumscribed  about  a 
sphere  whose  diameter  is  two-thirds  the  altitude  of  the 
cone.  Prove  that  its  lateral  surface  and  volume  are, 
respectively,  three-halves  and  nine-fourths  the  surface 
and  volume  of  the  sphere. 


55.  If  the  radius  of  a  sphere  is  25,  find  the  lateral 
area  and  volume  of  an  inscribed  cone,  the  radius  of     ^f^ 
whose  base  is  24. 

(Two  solutions.) 

56.  If  the  volume  of  a  sphere  is  ^^  ir,  find  the  lateral  area  and 
volume  of  a  circumscribed  cone  whose  altitude  is  18. 

57.  Find  the  volume  of  a  spherical  segment  of  one  base  whose 
altitude  is  6,  the  diameter  of  the  sphere  being  30. 

B 

58.  A  square  whose  area  is  A  revolves  about  its  diago- 
nal as  an  axis.     Find  the  area  of  the  entire  surface,  an;'  c 
the  volume,  of  the  solid  generated. 


59.  The  altitude  of  a  cone  of  revolution  is  9.  At  what  distances 
from  the  vertex  must  it  be  cut  by  planes  parallel  to  its  base,  in  order 
that  it  may  be  divided  into  three  equivalent  parts  ?     (§  656.) 

(Let  V  denote  the  volume  of  the  cone,  x  the  distance  from  the 
vertex  to  the  nearer  plane,  and  y  the  distance  to  the  other.) 

60.  Given  the  radius  of  the  base,  B,  and  the  total  area,  T,  of  a 
cylinder  of  revolution,  to  find  its  volume. 

(Find  H  from  the  equation  T-2irBH^2  irB^.) 

61.  Given  the  diameter  of  the  base,  D,  and  the  volume,  F,  of  a 
cylinder  of  revolution,  to  find  its  lateral  area  and  total  area. 

62.  Given  the  altitude,  H,  and  the  volume,  F,  of  a  cone  of  revo- 
lution, to  find  its  lateral  area. 

63.  Given  the  slant  height,  i,  and  the  lateral  area,  8,  of  a  cone 
of  revolution,  to  find  its  volume. 


MEASUREMENT  OF  THE   SPHERE. 


383 


64.  A  circular  sector  whose  central  angle  is  45''  and  radius  12 
revolves  about  a  diameter  perpendicular  to  one  of  its  bounding  radii. 
Find  the  volume  of  the  spherical  sector  generated. 

65.  Given  the  area  of  the  surface  of  a  sphere,  S,  to  find  its 
volume. 

66.  Given  the  volume  of  a  sphere,  F,  to  find  the  area  of  its 
surface. 

67.  A  right  triangle,  whose  legs  are  a  and  b,  revolves  about  its 
hypotenuse  as  an  axis.  Find  the  area  of  the  entire  surface,  and  the 
volume,  of  the  solid  generated. 


68.  The  parallel  sides  of  a  trapezoid  are  12  and  26, 
respectively,  and  its  non-parallel  sides  are  13  and  15. 
Find  the  volume  generated  by  the  revolution  of  the 
trapezoid  about  its  longest  side  as  an  axis. 

(Represent  BE  by  x.) 


69.  An  equilateral  triangle,  whose  altitude  is  A,  revolves  about  one 
of  its  altitudes  as  an  axis.  Find  the  area  of  the  surface,  and  the 
volume,  of  the  solids  generated  by  the  triangle,  and  by  its  inscribed 
circle.     (Ex.  21,  p.  151.) 

70.  Find  the  lateral  area  and  volume  of  a  cylinder  of  revolution, 
whose  altitude  is  equal  to  the  diameter  of  its  base,  inscribed  in  a  cone 
of  revolution  whose  altitude  is  h,  and  radius  of  base  r. 

(Represent  altitude  of  cylinder  by  x.) 

71.  Find  the  lateral  area  and  volume  of  a  cylinder  of  revolution, 
whose  altitude  is  equal  to  the  diameter  of  its  base,  inscribed  in  a 
sphere  whose  radius  is  r. 

72.  An  equilateral  triangle,  whose  side  is  a,  revolves 
about  a  straight  line  drawn  through  one  of  its  vertices 
parallel  to  the  opposite  side.  Find  the  area  of  the  en- 
tire surface,  and'the  volume,  of  the  solid  generated. 

(The  solid  generated  is  the  difference  of  the  cylinder 
generated  by  BCHG,  and  the  cones  generated  by  ABG 
and  ACH.) 

73.  The  outer  diameter  of  a  spherical  shell  is  9  in.,  and  its  thick- 
ness is  1  in.  What  is  its  weight,  if  a  cubic  inch  of  the  metal  weighs 
Jib.?     (ir  =  3.1416.) 


384 


SOLID  GEOMETRY.— BOOK  IX. 


74.  Find  the  diameter  of  a  sphere  in  which  the  area  of  the  sur- 
face and  the  volume  are  expressed  by  the  same  numbers. 

75.  A  regular  hexagon,  whose  side  is  «,  revolves  about  its  longest 
diagonal  as  an  axis.  Find  the  area  of  the  entire  surface,  and  the 
volume,  of  the  solid  generated. 

76.  The  sides  AB  and  BC  of  rectangle  ABCD  are  5  and  8,  respec- 
tively. Find  the  volumes  generated  by  the  revolution  of  triangle 
ACD  about  sides  AB  and  BC  as  axes. 

77.  The  sides  of  a  triangle  are  17,  25,  and  28.  Find  the  volume 
generated  by  the  revolution  of  the  triangle  about  its  longest  side  as 
an  axis.     (§  324.) 

78.  A  frustum  of  a  circular  cone  is  equivalent  to  three  cones, 
whose  common  altitude  is  the  altitude  of  the  frustum,  and  whose 
bases  are  the  lower  base,  the  upper  base,  and  a  mean  proportional 
between  the  bases  of  the  frustum.     (§  660.) 

79.  The  volume  of  a  cone  of  revolution  is  equal  to  the  area  of  its 
generating  triangle,  multiplied  by  the  circumference  of  a  circle  whose 
radius  is  the  distance  to  the  axis  from  the  intersection  of  the  medians 
of  the  triangle.     (§140.) 


80.  If  the  earth  be  refgarded  as  a  sphere  whose  radius 
is  B,  what  is  the  area  of  the  zone  visible  from  a  point 
whose  height  above  the  surface  is  if?     (§  271,  2.) 


81.  The  sides  AB  and  BC  of  acute-angled 
triangle  ABC  are  \/241  and  10,  respectively. 
Find  the  volume  of  the  solid  generated  by  the 
revolution  of  the  triangle  about  an  axis  in  its 
plane,  not  intersecting  its  surface,  whose  dis- 
tances from  A,  B,  and  C  are  2,  17,  and  11, 
respectively. 

82.  A  projectile  consists  of  two  hemispheres,  connected 
cylinder  of  revolution.  If  the  altitude  and  diameter  of  the  base 
cylinder  are  8  in.  and  7  in.,  respectively,  find  the  number  of 
inches  in  the  projectile,     (tt  =  3.1416.) 


83.  A  segment  of  a  circle,  whose  bounding  arc  is  a 
quadrant,  and  whose  radius  is  r,  revolves  about  a  diameter 
parallel  to  its  bounding  chord.  Find  the  area  of  the'entire 
surface,  and  the  volume,  of  the  solid  generated. 


by  a 
of  the 
cubic 


MEASUREMENT  OF  THE   SPHERE. 


385 


84.  If  any  triangle  be  revolved  about  an  axis  in  its  plane,  not 
parallel  to  its  base,  vsrhich  passes  through  its  vertex  without  intersect- 
ing its  surface,  the  volume  of  the  solid  generated  is  equal  to  the  area 
of  the  surface  generated  by  the  base,  multiplied  by  one-third  the 
altitude. 


Fig.  2. 


Fig.  S. 


(Compare  §  666.  Case  I.,  Figs.  1  and  2,  when  a  side  coincides  with 
the  axis ;  there  are  two  cases  according  as  AE  falls  on  BC,  or  BC 
produced.  Case  II. ,  Fig.  3,  when  no  side  coincides  with  the  axis ; 
prove  by  Case  I.) 

85.  If  any  triangle  be  revolved  about  an  axis  which  passes  through 
its  vertex  parallel  to  its  base,  the 
volume  of  the  solid  generated  is  equal 
to  the  area  of  the  surface  generated 
by  the  base,  multiplied  by  one-third 
the  altitude. 

(Compare  Ex.  72,  p.  383.  There 
are  two  cases  according  as  AD  falls 
on  5C,  OT  BC  produced.) 


86.  Find  the  area  of  the  surface  of  the 
sphere  circumscribing  a  regular  tetraedron, 
whose  edge  is  8. 

(Draw  lines  DOE  and  AOF  ±  to  A  ABC 
and  BCD,  respectively.) 


386  SOLID   GEOMETRY. 


APPENDIX. 

PROOF  OF   STATEMENT  MADE   IN  ELEVENTH  LINE, 
PAGE   201. 

683.  Theorem.  The  circumference  of  a  circle  is  shorter  than 
the  perimeter  of  any  circumscribed  polygon.  ''r-.^E 

Given  polygon  ABCD  circumscribed  about  a  O. 

To    Prove  circumference  of    O  shorter  than 
perimeter  ABCD. 

Proof.     Of  the  perimeters  of  the  O  and  of  its 
circumscribed  polygons,  there  must  be  one  perime-     A 
ter  such  that  all  the  others  are  of  equal  or  greater  length. 

But  no  circumscribed  polygon  can  have  this  perimeter. 

For,  if  we  suppose  polygon  ABCD  to  have  this  perimeter,  and  draw 
a  tangent  to  the  O,  meeting  CD  and  DA  at  points  E  and  F,  respec- 
tively, then  since  str.  line  EF  is  <  broken  line  EDF.,  the  perimeter  of 
circumscribed  polygon  ABCEF  is  <  perimeter  ABCD. 

Hence,  the  circumference  of  the  O  is  <  the  perimeter  of  any  cir- 
cumscribed polygon. 

PROOFS  OF  THE  LIMIT  STATEMENTS  OF   §640. 

684.  We  assume  the  following  : 

A  portion  of  a  plane  is  less  than  any  other  surface  having  the  same 
boundaries. 

685.  Theorem.  The  total  surface  of  a  circular  cylinder  is  less 
than  the  total  surface  of  any  circumscribed 

prism.* 

Given  prism  AC  circumscribed  about 
circular  cylinder  EG. 

To  Prove  total  surface  EGr  <  total  sur- 
face AC. 

Proof.  Of  the  total  surfaces  of  the 
cylinder  and  of  its  circumscribed  prisms, 
there  must  be  one  total  surface  such  that 
the  area  of  every  other  is  either  equal  to 
or  >  it. 


APPENDIX.  387 

But  no  circumscribed  prism  can  have  this  total  surface. 

For  suppose  prism  AC  to  have  this  total  surface;  and  let 
BGDFE  —  E'  be  a  circumscribed  prism,  whose  face  EF'  intersects 
faces  AB'  and  AD'  in  lines  EE'  and  FF',  respectively. 

Now,  face  EF'  is  <  sum  of  faces  AE',  AF',  AEF,  and  A'E'F'. 

(§684) 

Whence,  total  surface  of  prism  BCDFE  —  E'  is  <  total  surface  of 
prism  AC. 

Then,  total  surface  of  cylinder  EG  is  <  total  surface  of  any  cir- 
cumscribed prism. 

Proofs  of  the  Limit  Statements  of  §  640. 

686.  Let  L  denote  the  lateral  edge,  H  the  altitude,  S  and  s  the 
the  lateral  areas,  V  and  v  the  volumes,  E  and  e  the  perimeters  of  rt. 
sections,  and  B  and  6  the  areas  of  the  bases  of  the  circumscribed  and 
inscribed  prisms,  respectively  ;  also,  JS'  the  lateral  area  of  the  cylinder, 
V  its  volume,  E'  the  perimeter  of  a  rt.  section,  and  B'  the  area  of  the 


1.   We  have,  8 +  2  B>  S'  +  2  B'.  (§685) 

.-.  S+2(B-  B>)  >  S'. 

Again,  the  total  surface  of  the  inscribed  prism  is  <  the  total  surface 
of  the  cylinder.  (§  684) 

.-.  S<  -\-2B'>s-{-2h,   or   S'>s^2{h-  B'). 

Then,  S+2(B-B')>S'>s  +  2(b-B'). 

Now  if  the  number  of  faces  of  the  prisms  be  indefinitely  increased, 
B-  B'  andb  -  B'  approach  the  limit  0.  (§  363,  II) 

Again,  the  difference  between  the  perimeters  of  the  bases  of  the 
prisms  approaches  the  limit  0.  (§  363,  I) 

Then,  the  total  surface  of  the  circumscribed  prism  continually  de- 
creases, but  never  reaches  the  total  surface  of  the  inscribed  prism  ; 
and  the  total  surface  of  the  inscribed  prism  continually  increases,  but 
never  reaches  the  total  surface  of  the  circumscribed  prism.       (§  684) 

Then,  the  difference  between  S  +  2  B  and  s  +  2b  can  be  made  less 
than  any  assigned  value,  however  small. 

Whence,  S  +  2  B  -  (s  -\-  2  b) ,  or  S  -  s  -\-  2  (B  -  b),  approaches  the 
limit  0. 

But  B  —  b  approaches  the  limit  0.  (§  363,  II) 

Whence,  S  —  s  approaches  the  limit  0. 

Then,  S'  is  intermediate  in  value  between  two  variables,  the  differ- 
ence between  which  approaches  the  limit  0. 


388  SOLID   GEOMETRY. 

Then,  the  difference  between  either  variable  and  S',  that  is, 
S  +  2(B  -B>)  -  S'  and  S'  -s-2(b-  B'), 
approaches  the  limit  0. 

Whence,  S  —  S'  and  S'  —  s  approach  the  limit  0. 
Hence,  S  and  s  approach  the  limit  S'. 

2.  We  have,  V=  B  x  H  &ndv  =  b  x  H.  (§  499) 
Whence,       V-v  =  BxH-bx  H=  (B  -  b)  x  H. 

Now  if  the  number  of  faces  of  the  prisms  be  indefinitely  increased, 
B  —  b,  and  therefore  V—  v,  approaches  the  limit  0.  (§  363,  II) 

But  V  is  evidently  >  v,  and  <  V. 
Then,  V—  V  and  V  —  v  approach  the  limit  0. 
Whence,  V  and  v  approach  the  limit  V. 

3.  We  have,  S  =  E  x  L  and  s  =  e  x  L.  (§  484) 

Then,  ^  =  -  and  e=-;  OT,U-e  =  ^=-?. 

L  L  L 

Now  if  the  number  of  faces  of  the  prisms  be  indefinitely  increased, 
8  —  s^  and  therefore  E  —  e,  approaches  the  limit  0.  (§  640,  1) 

But  E',  the  perimeter  of  a  rt.  section  of  the  cylinder,  is  <  JE";  for 
the  theorem  of  §  683  is  evidently  true  when  for  the  O  is  taken  any 
closed  curve  whose  tangents  do  not  intersect  its  surface  ;  also,  E' 
is  >  e.  (Ax.  4) 

Then,  E  —  E'  and  E'  —  e  approach  the  limit  0. 

Whence,  E  and  e  approach  the  limit  E'. 

PROOFS  OF  THE   LIMIT  STATEMENTS  OF   §650. 

687.  Theorem.  The  total  surface  of  a  circular  cone  is  less  than 
the  total  surface  of  any  circumscribed  pyramid. 

Given  pyramid  S-ABCD  circumscribed 
about  circular  cone  S-EF. 

To  Prove  total  surface  S-EF<C  total  sur- 
face S-ABCD. 

Proof.  Of  the  total  surfaces  of  the  cone 
and  of  its  circumscribed  pyramids,  there  must 
be  one  total  surface  such  that  the  area  of  every 
other  is  either  equal  to  or  >  it. 

But  no  circumscribed  pyramid  can  have  this  J,otal  surface. 

For  suppose  pyramid  S-ABCD  to  have  this  total  surface ;  and  let 
S-BCDFE  be  a  circumscribed  pyramid,  whose  face  SEF  intersects 
SAB  and  SAD  in  lines  ^S*^  and  SF,  respectively. 


APPENDIX.  389 

Now,  face  SEF  is  <  sum  of  faces  8AE,  SAF,  and  AEF.    (§  684) 
Whence,  total  surface  of  pyramid  S-BCDFE  is  <  total  surface 
of  pyramid  S-ABCD. 

Then,  total  surface  of  cone  S-EF  is  <  total  surface  of  any  circum- 
scribed pyramid. 

Proofs  of  the  Limit  Statements  of  §  650. 

688.  Let  H  denote  the  altitude,  S  and  s  the  lateral  areas,  V  and 
V  the  volumes,  and  B  and  h  the  areas  of  the  bases,  of  the  circum- 
scribed and  inscribed  pyramids,  respectively  ;  also,  S'  the  lateral  area 
of  the  cone,  V  its  volume,  and  B'  the  area  of  its  base. 

1.  We  have,  S-\-  B>8'  +  B<.  (§  687) 

.-.  ^4-  {B-B')>S'. 

Again,  the  total  surface  of  the  inscribed  pyramid  is  <  the  total 
surface  of  the  cone.  (§  684) 

.-.  8'  +  B<>s^-h,  or  S'>s^{h-B'). 

Then,  8 -\- {B -  Bi)>  8' > s -\- {h  -  B>). 

Now  if  the  number  of  faces  of  the  pyramids  be  indefinitely  in- 
creased, 5  -  5'  and  6  -  B'  approach  the  limit  0.  (§  363,  II) 

Also,  the  difference  between  the  perimeters  of  the  bases  of  the 
pyramids  approaches  the  limit  0.  (§  363,  I) 

Then,  8 -V  B  continually  decreases,  and  s  -\- h  continually  increases ; 
and  the  difference  between  them  can  be  made  less  than  any  assigned 
value,  however  small.  (§  684) 

Then,  8-  s-\-{B  -h)  approaches  the  limit  0. 

But  5  -  &  approaches  the  limit  0.  (§  363,  II) 

Whence,  8  —  s  approaches  the  limit  0. 

Then,  8'  is  intermediate  in  value  between  two  variables,  the  differ- 
ence between  which  approaches  the  limit  0. 

Whence,  the  difference  between  either  variable  and  8',  that  is, 
8-\-{B-B')-  8'  and  8'  -s-(b-B'),  approaches  the  limit  0. 

Then,  8—8'  and  8'  —  s  approach  the  limit  0. 

Whence,  8  and  s  approach  the  limit  8'. 

2.  We  have,  V=  B  x  }  H  Rnd  v  =  h  x  ^  H.  (§  521) 
Whence,                  V  -  v  =  {B  -  b)  x  ^  H. 

Now  if  the  number  of  faces  of  the  pyramids  be  indefinitely  increased, 
B  -b,  and  therefore  V—v,  approaches  the  limit  0.  (§  363,  II) 

But,  V  is  evidently  >  v,  and  <  V. 
Then,  V-V  and  V  -  v  approach  the  limit  0. 
Whence,  V  and  v  approach  the  limit  V'. 


390  SOLID   GEOMETRY. 


PROOF  OF  THE   LIMIT  STATEMENT  IN   NOTE   FOOT 
OF  PAGE   374. 

689.  Theorem.  If  a  regular  broken  liiie,  inscribed  in  an  arc. 
be  revolved  about  a  diameter,  not  intersecting  the 
arc,  as  an  axis,  and  the  subdivisions  of  the  arc 
be  bisected  indefinitely,  the  area  of  the  surface 
generated  by  the  broken  line  approaches  the  area 
of  the  surface  generated  by  the  arc  as  a  limit. 

Given  regular  broken  line  ABCD,  inscribed  in 
arc  AD,  revolving  about  diameter  OM  as  an  axis. 

To  Prove  that,  if  the  subdivisions  of  arc  AD 
be  bisected  indefinitely,  area  of  surface  generated 
by  ABCD  approaches  area  of  surface  generated  by 
arc  AD  as  a  limit. 

Proof.  Let  A'B',  B'C,  and  CD'  be  tangents  ||  to  AB,  BC,  and 
CD,  respectively,  points  A',  B',  C,  and  D'  being  in  radii  OA,  OB, 
OC,  and  OD,  respectively,  produced  ;  and  let  S,  s,  and  S'  denote  the 
areas  of  the  surfaces  generated  by  A'B' CD',  and  ABCD,  and  arc 
AD,  respectively. 

Of  the  surfaces  generated  by  arc  AD,  by  ABCD,  and  by  regular 
inscribed  broken  lines  obtained  by  bisecting  the  subdivisions  of  the 
arc  indefinitely,  there  must  be  one  surface  such  that  the  areas  of  all 
the  others  are  either  equal  to  or  <  it. 

But  no  regular  inscribed  broken  line  can  generate  this  surface. 

For  if  this  were  the  case,  by  bisecting  the  subdivisions  of  the  arc,  a 
regular  inscribed  broken  line  would  be  obtained  having  the  same  pro- 
jection on  the  axis ;  but  the  ±  from  0  to  each  line  would  be  greater, 
and  hence  the  surface  generated  would  be  greater. 

(§  665,  and  Note  foot  of  p.  374.) 

Hence,  surface  generated  by  arc  AD  is  >  surface  generated  by 
ABCD  ;  that  is,  S'  is  >  s. 

Again,  of  the  surfaces  generated  by  arc  AD,  by  A'B' CD',  and  by 
regular  circumscribed  broken  lines  obtained  by  bisecting  the  sub- 
divisions of  the  arc  indefinitely,  there  must  be  one  surface  such  that 
the  areas  of  all  the  others  are  either  equal  to  or  >  it. 

But  po  regular  circumscribed  broken  line  can  generate  this  surface. 

For  if  this  were  the  case,  by  bisecting  the  subdivisions  of  the  arc,  a 
regular  circumscribed  broken  line  would  be  obtained  in  which  the  _L 
from  0  to  each  line  would  be  the  same  ;  but  the  projection  on  the  axis 
would  be  smaller,  and  hence  tihp  ;sH.rf.ape  generated  would  be  smaller. 


APPENDIX.  391 

Hence,  surface  generated  by  arc  AD  is  <  surface  generated  by 
A'B'C'D' ;  that  is,  >S"  is  <  S. 

Then,  S  -  S'  and  S'  -  s  are  <  S  -  s. 

Now  if  the  subdivisions  of  arc  AD  be  bisected  indefinitely,  the 
difference  between  broken  lines  A'B'C'D'  and  ABCD  approaches  the 
limit  0.     (Note  foot  p.  374.) 

Then,  the  difference  between  the  projections  on  OM  oi  A'B'C'D' 
and  ABCD  approaches  the  limit  0. 

Also,  the  difference  between  the  Js  from  O  to  A'B'  and  AB  ap- 
proaches the  limit  0.     (Note  foot  p.  374. ) 

Then,  the  difference  between  the  areas  of  the  surfaces  generated 
by  A'B'C'D'  and  ABCD,  that  is,  8-s  approaches  the  limit  0.  (§  665) 

Then,  S  —  S'  and  S'  —  s  approach  the  limit  0. 

Whence,  S  and  s  approach  the  limit  ^S^'. 


PROOF  OF  THE  LIMIT  STATEMENT  IN  NOTE  FOOT 
OF  PAGE  377. 

690.  Theorem.  If  a  regular  polygonal  sector,  inscribed  in  a 
sector  of  a  circle,  be  revolved  about  a  diameter,  not  crossing  the  sector, 
as  an  axis,  and  the  subdivisions  of  the  arc  be  bisected  indefinitely,  the 
volume  of  the  solid  generated  by  the  polygonal  sector  approaches  the 
volume  of  the  solid  generated  by  the  sector  as  a  limit. 

Given  regular  polygonal  sector  OABCD,  inscribed  in  sector  OAD, 
revolved  about  diameter  OM  as  an  axis.     (Fig.  of  §  689. ) 

To  Prove  that,  if  the  subdivisions  of  arc  AD  be  bisected  indefi- 
nitely, volume  of  solid  generated  by  OABCD  approaches  volume  of 
solid  generated  by  sector  OAD  as  a  limit. 

Proof.  Let  A'B',  B'C,  and  CD'  be  tangents  ||  to  AB,  BC,  and 
CD,  respectively,  points  A',  B',  C,  and  D'  being  in  radii  OA,  OB, 
OC,  and  OD,  respectively,  produced  ;  and  let  V,  v,  and  V  denote 
the  volumes  of  the  solids  generated  by  OA'B'C'D',  OABCD,  and 
sector  OAD,  respectively. 

Then,  V  is  evidently  >  v,  and  <  F. 

Whence,  V  —  V  and  V  —  v  are  <  F—  v. 

Now  if  the  subdivisions  of  arc  AD  be  biseqted  indefinitely,  the 
difference  between  the  areas  of  OA'B'C'D'  and  OABCD,  and  there- 
fore V—  v,  approaches  the  limit  0.     (Note  foot  p.  377.) 

Then,  V  —  V  and  V  —  v  approach  the  limit  0. 

Whence,  V  and  v  approach  the  limit  V. 


SUGGESTIONS  FOR  THE  USE  OF  COLORED 
PLATES. 

In  beginning  the  study  of  solid  geometry,  a  new  difficulty 
is  encountered,  the  difficulty  of  seeing  the  figures  correctly. 
Through  plane  geometry,  the  pupil  has  acquired  the  habit 
of  looking  at  the  figures  simply  as  lines  making  different 
angles  and  running  in  varying  directions,  but  always  limited 
to  one  plane.  To  the  untrained  eye,  the  line  figures  in 
solid  geometry  do  not  look  essentially  different.  The 
teacher  sees,  the  pupil  does  not,  and,  worse  than  all,  too 
frequently  the  teacher  fails  to  realize  that  what  represents 
to  him  a  solid  figure  is,  to  the  pupil,  a  number  of  lines 
similar  to  those  in  plane  geometry,  only  hopelessly  compli- 
cated in  arrangement. 

The  first  thing  necessary,  then,  is  to  train  a  class  to 
visualize  correctly,  —  to  see  in  imagination,  not  a  seeming 
confusion  of  lines,  but  the  solids  outlined  by  those  lines. 

In  the  hope  of  accomplishing  this,  various  aids  have  been 
offered  by  text-books  in  the  way  of  graphic  representation, 
but  all  of  them,  while  attractive  at  first,  have,  when  tried, 
fallen  short  of  expectation  in  teaching  value. 

Work  with  actual  models  is  accurate  and  helpful,  but  pho- 
tographic reproductions  of  these  models  make  nearly  the 
same  demands  upon  the  untrained  imagination  that  the  line 
figures  do.  Shaded  figures  have  been  used,  but  the  simi- 
larity in  tone  of  grays  and  blacks  is  confusing  to  the  unedu- 
cated eye. 

With  the   color  scheme  here   presented,  the   confusion 


394  SOLID   GEOMETRY. 

vanishes,  and  the  pupil  not  only  may  see,  but  must  see,  the 
planes  in  their  true  relations  to  each  other.  If  his  first 
glimpse  of  figures  for  solids  is  right,  he  is  ready  then  to 
look  for  depth,  distance,  and  three  dimensions,  in  all  suc- 
ceeding figures. 

The  few  colored  figures  here  presented  are  valuable  in  the 
beginning,  to  show  the  pupil  the  kind  of  thing  that  he  is  to 
look  for  —  what  he  is  expected  to  see.  Take,  for  instance, 
the  figure  on  page  236.  To  the  beginner  there  is  little  sug- 
gestion of  various  planes  intersecting,  disappearing  behind 
each  other,  and  reappearing,  but  by  Plate  I  all  this  is  in- 
stantly revealed.  The  correct  visual  impression  here  gained 
will  then  be  transferred  naturally  to  the  line  figure. 

Another  objection  to  the  aids  thus  far  presented  lies  in 
the  fact  that  the  text-book  does  all  the  work,  leaving  the 
pupil  only  an  observer.  If  the  work  stops  with  looking  at 
the  figures  and  studying  from  them,  their  greatest  teaching 
value  is  lost.  It  is  comparatively  easy,  with  a  figure  that 
has  been  carefully  drawn  and  effectively  shaded  or  colored, 
to  grasp  for  the  moment  the  general  idea  indicated,  but  the 
impression  will  be  neither  complete  nor  lasting. 

Purposely  only  a  few  suggestive  figures  are  here  pre- 
sented in  color,  it  being  the  plan  that  the  pupil,  from  the 
figures  given  in  the  text,  and  from  the  accompanying  dem- 
onstration, shall  interpret  in  color  the  solids  indicated. 
When  he  is  compelled  thus  to  fix  the  limitations  of  the 
planes,  he  is  led  to  definite  knowledge  that  is  otherwise 
impossible.  Here  is  represented  all  the  vast  distance  that 
lies  between  looking  at  a  picture  done  by  some  one  else,  and 
reproducing  that  picture  yourself,  —  all  the  difference  be- 
tween observing  and  doing. 

The  plan  here  offered  is  capable  of  practical  application 
in  several  ways. 

Send  the  class  to  the  board  to  draw  the  figures  for  the 
day  with  colored  crayons ;  the  result  will  reveal  their  under- 
standing or  misunderstanding  of  the  proposition  under  con- 


THE  USE  OF  COLORED  PLATES.        395 

sideration.  With  the  color  in  their  own  hands,  pupils  are 
compelled  to  decide  where  plane  intersects  plane,  where  one 
disappears  behind  another,  and  many  other  things  that 
escaped  their  observation  in  studying  from  the  book.  Take, 
for  instance,  the  figure  on  page  236.  It  probably  never 
occurred  to  the  pupil  in  studying  to  observe  into  how  many 
planes  the  argument  is  carried.  When  he  colors  it  he  must 
know. 

This  first  work  should  be  done  rapidly,  with  no  attempt 
at  finished  drawings.  Sometimes  it  is  well  to  have  a  class 
draw  entirely  free-hand,  laying  in  the  color  rapidly,  attempt- 
ing only  to  bring  out  the  geometric  idea.  At  first  strongly 
contrasting  colors  should  be  used,  and,  as  the  work  is  not 
permanent,  they  may  be  even  crude,  if  only  striking. 

Following  this,  certain  figures  should  be  put  into  perma- 
nent form.  Such  drawing  should  be  done  carefully,  with  as 
close  mathematical  accuracy  as  board  and  chalk  will  allow. 
Here  attention  should  be  given  to  the  color  scheme  and  the 
result  made  restful  to  the  eye.  There  is  actual  teaching  value 
in  having  these  difficult  figures  long  before  the  attention. 

With  the  figures  thus  before  the  class,  it  is  easy  in  the 
few  spare  moments  that  occasionally  come  at  the  close  of  a 
recitation,  to  give  a  quicks  review  that  would  not  be  possible 
if  time  had  to  be  consumed  in  drawing. 

Blackboard  work  is  strengthened  by  outlining  all  planes 
in  strong  white  lines,  just  as  in  the  book  they  are  outlined 
in  black. 

Another  useful  expedient  in  the  use  of  color  is  the  careful 
preparation  of  plates  outside  of  class.  The  most  interesting 
and  effective  figures  should  be  selected,  and  all  members  of 
the  class  required  to  execute  a  certain  number,  this  number 
varying  according  to  the  ability  of  different  classes.  It  is 
usually  well  to  suggest  a  uniform  size  of  paper ;  seven  by 
nine  inches,  approximately,  is  desirable.  One  figure  only 
should  be  placed  on  a  sheet.  As  to  size  of  figure,  it  is  bet- 
ter not  to  dictate.     When  the  first  drawings  are  brought 


393  SOLID   GEOMETRY. 

together,  it  will  not  take  a  class  long  to  decide  which  is 
most  effective.  From  this  they  will  modify  their  scale, 
approximating  the  best,  but  retaining  perfect  individuality. 
.  One  additional  direction  should  be  given,  applying  equally 
to  board  work  and  to  work  on  paper.  Leave  only  such  lines 
as  would  be  visible  if  the  planes  were  opaque.  A  glance  at 
the  colored  plates  here  given  will  show  that  by  omitting  the 
dotted  lines  the  figure  is  more  effective,  and  the  solidarity 
greatly  emphasized.  The  geometric  value  of  the  lines  is 
not  lost,  for  the  eye  naturally  carries  them  along  behind  the 
plane,  and  joins  the  parts  correctly  if  they  reappear.  In 
outlining  the  figure  at  first,  these  lines,  of  course,  should 
be  drawn,  for  by  them  is  frequently  determined  where  the 
planes  intersect. 

As  to  the  medium  used,  that  is  a  matter  of  taste  and 
equipment.  Colored  pencils  are  easiest  for  the  untrained 
hand,  and  good  effects  can  be  obtained  with  them.  If  any  one 
in  the  class  handles  water  colors,  he  should  be  encouraged 
to  use  them,  for  they  make  stronger  figures,  and  the  influ- 
ence of  even  one  or  two  working  in  them  will  elevate  the 
entire  standard. 

Some  or  all  of  these  expedients  may  be  used  as  the  condi- 
tions of  individual  classes  indicate,  but  let  it  always  be 
insisted  that  the  class  do  the  work.  The  most  carefully 
executed  drawing  of  teacher  or  text-book  is  worth  less  than 
the  poorest  attempt  of  the  poorest  pupil. 

Finally,  the  method  here  presented  is  offered  only  as  a 
practical  suggestion  for  clearer  teaching,  not  as  an  integral 
part  of  geometry,  and  may  be  used  or  not  as  teachers  desire. 
Like  everything  else,  it  is  capable  of  abuse  and  perversion, 
and  whoever  uses  it  should  be  ever  watchful  lest  it  overstep 
its  proper  limitations.  Its  purpose  is  not  to  produce  a  fine 
set  of  drawings,  but  to  assist  in  teaching,  geometry.  It  is  a 
means,  not  an  end ;  an  expedient,  not  a  science. 


Plate  I. 


Plate  II. 


Plate  III. 


Plate  IV. 


Plate  VI. 


Plate  VIII. 


/  (M 

/ 

/<^ 

/ 

/ 

Plate  IX. 


Plate 


INDEX  TO   DEFINITIONS. 


Acute  angle,  §  27. 
Adjacent  angles,  §  23. 

diedral  angles,  §  428. 
Alternate-exterior  angles,  §  71. 
-interior  angles,  §  71. 
Alternation,  §  235. 
Altitude  of  a  cone,  §  553. 

of  a  cylinder,  §  540. 

of  a  frustum  of  a  cone, 

§  553. 
of  a  frustum  of  a  pyramid, 

§606. 
of  a  parallelogram,  ^  104. 
of  a  prism,  §  466. 
of  a  pyramid,  §  502. 
of  a  spherical   segment, 

§662. 
of  a  trapezoid,  §  104. 
of  a  triangle,  §  60. 
of  a  zone,  §  662. 
Angle,  §  20. 

at  the  centre  of  a  regular 

polygon,  §  341. 
between    two    intersecting 

curves,  §  583. 
inscribed    in    a    segment, 

§148. 
of  a  lune,  §  616. 
Angles  of  a  polygon,  §  118. 

of  a  quadrilateral,  §  103. 
of    a    spherical    polygon, 

§  587. 
of  a  triangle,  §  57. 
Angular  degree,  §  29. 
Antecedents  of  a  proportion,  §  229. 


Apothem  of   a  regular   polygon, 

§  341. 
Arc  of  a  circle,  §  142. 
Area  of  a  surface,  §  302. 
Axiom,  §  15. 

Axis  of  a  circle  of  a  sphere,  §  567. 
of  a  circular  cone,  §  553. 
of  a  circular  cylinder,  §  546. 
•     of  symmetry,  §  387. 

Base  of  a  cone,  §  553. 

of  a  polyedral  angle,  §  452. 
of  a  pyramid,  §  502. 
of  a  spherical  pyramid,  §  589. 
of  a  spherical  sector,  §  664. 
of  a  spherical  wedge,  §  617. 
of  a  triangle,  §  60. 
Bases  of  a  cylinder,  §  540. 

of  a  parallelogram,  §  104. 

of  a  prism,  §  466. 

of  a  spherical  segment,  §  662. 

of  a  trapezoid,  §  104. 

of  a  truncated  prism,  §  472. 

of    a    truncated    pyramid, 

§  505. 
of  a  zone,  §  662. 
Bi- rectangular  triangle,  §  598. 
Broken  line,  §  7. 

Central  angle,  §  148. 
Centre  of  a  circle,  §  142. 

of  a  parallelogram,  §  111, 

of  a  regular  polygon,  §  341 

of  a  sphere,  §  561. 

of  symmetry,  §  386. 


397 


398 


INDEX  TO   DEFINITIONS. 


Chord  of  a  circle,  §  147. 
Circle,  §  142. 

circumscribed  about  a  poly- 
gon, §  151. 
inscribed    in    a    polygon, 
§151. 
Circles  tangent  externally,  §  150. 
tangent  internally,  §  150. 
Circular  cone,  §  553. 

cylinder,  §  540. 
Circumference,  §  142. 
Commensurable  magnitudes,  §  181. 
Common  measure,  §  181. 
tangent,  §  150. 
Complement  of  an  angle,  §  30. 
of  an  arc,  §  190. 
Complementary  angles,  §  30. 
Composition,  §  237. 
Concave  polygon,  §  121. 
Concentric  circles,  §  146. 
Conclusion,  §  38. 
Cone,  §  553. 

of  revolution,  §  555. 
Conical  surface,  §  553. 
Consequents  of  a  proportion,  §  229. 
Constant,  §  185. 
Converse  of  a  proposition,  §  39. 
Convex  polyedral  angle,  §  453. 
polyedron,  §  463. 
polygon,  §  121. 
spherical  polygon,  §  588. 
Corollary,  §  15. 
Corresponding  angles,  §  71. 
Cube,  §  474. 
Curve,  §  7. 
Curved  surface,  §  10. 
Cylinder,  §  640. 

of  revolution,  §  550. 
Cylindrical  surface,  §  540. 

Decagon,  §  119. 
Degree  of  arc,  §  190. 
Determination  of  a  plane,  §  394. 
of  a  straight  line, 
§18. 
Diagonal  of  a  polyedron,  §  461. 


Diagonal  of  a  polygon,  §  118. 

of  a  quadrilateral,  §  103. 
of  a  spherical   polygon, 
§587. 
Diameter  of  a  circle,  §  142. 
of  a  sphere,  §  561. 
Diedral  angle,  §  428. 
Dimensions  of  a  rectangle,  §  304. 
of  a  rectangular  paral- 
lelepiped, §  487. 
Directrix    of    a    conical    surface, 
§  553. 
of  a  cylindrical  surface, 
§540. 
Distance  between  two  points  on 
the  surface  of  a  sphere, 
§573. 
of  a  point  from  a  line, 

§47. 
of  a  point  from  a  plane, 
§410. 
Division,  §  238. 
Dodecaedron,  §  462. 
Dodecagon,  §  119. 

Edge  of  a  diedral  angle,  §  428. 
Edges  of  a  polyedral  angle,  §  452. 

of  a  polyedron,  §  461. 
Element  of  a  conical  surface,  §  553. 
of  a  cylindrical  surface, 
§540. 
Enneagon,  §  119. 
Equal  angles,  §  22. 

diedral  angles,  §  432. 
figures,  §  22. 
polyedral  angles,  §  454. 
Equiangular  polygon,  §  120. 

triangle,  §  58. 
Equilateral  polygon,  §  120. 

spherical  triangle, 

§587. 
triangle,  §  58. 
Equivalent  solids,  §  465. 

surfaces,  §  303. 
Exterior  angles,  §  71. 

of  a  triangle,  §  57. 


INDEX  TO   DEFINITIONS. 


399 


Extremes  of  a  proportion,  §  229. 

Face  angles  of  a  polyedral  angle, 

§  452. 
Faces  of  a  diedral  angle,  §  428. 
of  a  polyedral  angle,  §  452. 
of  a  polyedron,  §  461. 
Figure  symmetrical   with  respect 
to  a  centre,  §  390. 
symmetrical  with  respect 
to  an  axis,  §  391. 
Figures  symmetrical  with  respect 
to  a  centre,  §  388. 
symmetrical  with  respect 
to  an  axis,  §  388. 
Foot  of  a  line,  §  397. 
Fourth  proportional,  §  231. 
Frustum  of  a  cone,  §  653. 

of  a  pyramid,  §  506. 
of  a  pyramid  circum- 
scribed about  a  frus- 
tum of  a  cone,  §  649. 
of  a  pyramid  inscribed 
in  a  frustum  of  a  cone, 
§649. 

Greneratrix  of   a  conical  surface, 
§553. 
of   a   cylindrical  sur- 
face, §  540. 

Geometrical  figure,  §  12. 

Geometry,  §  13. 

Great  circle  of  a  sphere,  §  567. 

Hendecagon,  §  119. 

Heptagon,  §  119. 

Hexaedron,  §  462. 

Hexagon,  §  119. 

Homologous,  §§  65,  123. 

Hypotenuse  of    a  right  triangle, 

§59. 
Hypothesis,  §  15. 

Icosaedron,  §  462. 
Incommensurable         magnitudes, 
§181. 


Indirect  method  of  proof,  §  50. 
Inscribed  angle,  §  148. 
Inscriptible  polygon,  §  151. 
Interior  angles,  §  71. 
Inversion,  §  236. 
Isoperimetric  figures,  §  378. 
Isosceles  spherical  triangle,  §  587. 
triangle,  §  58. 

Lateral  area  of  a  cone,  §  647. 

of  a  cylinder,  §  638. 
of  a  frustum  of  a  cone, 

§647. 
of  a  prism,  §  466. 
of  a  pyramid,  §  502. 
Lateral  edges  of  a  prism,  §  466. 

of  a  pyramid,  §  502. 
Lateral  faces  of  a  prism,  §  466. 

of  a  pyramid,  §  502. 
Lateral  surface  of  a  cone,  §  553. 

of  a  cylinder,  §  540. 
Legs  of  a  right  triangle,  §  59. 
Limit  of  a  variable,  §  186. 
Line,  §  3. 

Locus  of  a  series  of  points,  §  141. 
Lower  base  of  a  frustum  of  a  cone, 
§  553. 
nappe  of  a  conical  surface, 
§553. 
Lune,  §  616. 


Material  body,  §  1. 
Mean  proportional,  §  230. 
Means  of  a  proportion,  §  229. 
Measure  of  a  magnitude,  §  180. 

of  an  angle,  §  29. 
Median  of  a  triangle,  §  139. 
Mutually    equiangular    polygons, 
§122. 
equiangular     spherical 

polygons,  §  599. 
equilateral      polygons, 

§  122. 
equilateral       spherical 
polygons,  §  599. 


400 


INDEX   TO   DEFINITIONS. 


Numerical  measure,  §  180. 

Oblique  angles,  §  27. 

lines,  §  27. 

prism,  §  470. 
Obtuse  angle,  §  27. 
Octaedron,  §  462. 
Octagon,  §  119. 

Parallel  lines,  §  52. 

planes,  §  397. 
Parallelogram,  §  104. 
Parallelopiped,  §  474. 
Pentagon,  §  119. 
Perimeter  of  a  polygon,  §  118. 
Perpendicular  Ijnes,  §  24. 

planes,  §  436. 
Plane,  §  9. 

angle  of  a  diedral    angle, 

§  429. 
figure,  §  12. 
geometry,  §  14. 
tangent  to  a  cone,  §  553. 
tangent  to  a  cylinder,  §  540. 
tangent  to  a  frustum  of  a 

cone,  §  553. 
tangent  to  a  sphere,  §  564. 
Point,  §  4. 

of  contact  of  a  line  tangent 

to  a  circle,  §  149. 
of  contact  of  a  line  tangent 

to  a  sphere,  §  564. 
of  contact  of  a  plane  tangent 
to  a  sphere,  §  564. 
Points  symmetrical  with  respect  to 
a  line,  §  387. 
symmetrical  with  respect  to 
a  point,  §  386. 
Polar  distance  of    a   circle  of    a 
sphere,  §  576.  ' 

triangle  of  a  spherical  tri- 
angle, §  590. 
triangles,  §  592. 
Poles  of  a  circle  of  a  sphere,  §  567. 
Polyedral  angle,  §  452. 
Polyedron,  §  461. 


Polyedron  circumscribed  about  a 
sphere,  §  564. 
inscribed  in  a  sphere, 
§564. 
Polygon,  §  118. 

circumscribed     about    a 

circle,  §  151. 
inscribed  in  a  circle,  §  151. 
Postulate,  §  15. 
Prism,  §  466. 

circumscribed  about  a  cylin- 
der, §  639. 
inscribed    in    a    cylinder, 
§  639. 
Problem,  §  15. 

Projection  of   a  line  on   a   line, 
§  276. 
of  a  line  on  a  plane, 

§  447. 
of  a  point  on  a  line, 

§  275. 
of  a  point  on  a  plane, 
§447. 
Proportion,  §  227. 
Proposition,  §  15. 
Pyramid,  §  502. 

circumscribed    about    a 

cone,  §  648. 
inscribed  in  a  cone,  §  648. 

Quadrangular  prism,  §  469. 

pyramid,  §  603. 
Quadrant,  §  146. 
Quadrilateral,  §  103. 

Kadius  of  a  circle,  §  142. 

of  a  regular  polygon,  §  341. 

of  a  sphere,  §  661. 
Ratio,  §  180. 

Reciprocally  proportional  magni- 
tudes,' §  281. 
Rectangle,  §  lj)5. 

Rectangular  parallelopiped,  §  474 
Rectilinear  figure,  §  12. 
Re-entrant  angle,  §  121. 
Regular  polyedron,  §  536. 


INDEX  TO  DEFINITIONS. 


403 


Regular  polygon,  §  339. 
prism,  §  471. 
pyramid,  §  504. 
Rhomboid,  §  105. 
Rhombus,  §  105. 
Right  angle,  §  24. 

angled    spherical    triangle, 
§  587. 

circular  cone,  §  553. 

cylinder,  §  540. 

diedral  angle,  §  436. 

parallelopiped,  §  474. 

prism,  §  470. 

section  of  a  cylinder,  §  638. 

section  of  a  prism,  §  473. 

triangle,  §  69. 

Scalene  triangle,  §  58. 
Scholium,  §  15. 
Secant,  §  149. 
Sector  of  a  circle,  §  147. 
Segment  of  a  circle,  §  147. 
Segments  of  a  line  by  a  point,  §  250. 
Semicircle,  §  147. 
Semi-circumference,  §  146. 
Sides  of  a  polygon,  §  118. 

of  a  quadrilateral,  §  103. 
of  a  spherical  polygon,  §  587. 
of  a  triangle,  §  57. 
of  an  angle,  §  20. 
Similar  arcs,  §  369. 

cones  of  revolution,  §  555. 
cylinders    of     revolution, 

§550. 
polyedrons,  §  527. 
polygons,  §  252. 
sectors,  §  369. 
segments,  §  369. 
Slant  height  of  a  cone  of  revolu- 
tion, §  647. 
of  a  frustum  of  a  cone 
ofrevolution,§647. 
of  a  frustum  of  a  regu- 
lar pyramid,  §  511. 
of  a  regular  pyramid, 
§508. 


Small  circle  of  a  sphere,  §  567. 
Solid,  §  2. 

geometry,  §  14. 
Sphere,  §  561. 

circumscribed      about     a 

polyedron,  §  564. 
inscribed  in  a  polyedron, 
§564. 
Spherical  angle,  §  583. 

excess  of  a  spherical  tri- 
angle, §  632. 
polygon,  §  587. 
pyramid,  §  589. 
sector,  §  664. 
segment,  §  662. 
segment    of    one    base, 

§  662. 
triangle,  §  587. 
wedge,  §  617. 
Square,  §  105. 
Straight  line,  §  7. 

divided  in  extreme 
and  mean  ratio 
externally,  §  296. 
divided  in  extreme 
and  mean  ratio 
internally,  §  296. 
oblique  to  a  plane, 

§397. 
parallel  to  a  plane, 

§  397. 
perpendicular    to    a 

plane,  §  397. 
tangent  to  a  circle, 

§  149. 
tangent  to  a  sphere, 
§  564. 
Straight  lines  divided  proportion- 
ally, §  243. 
Sublended  arc,  §  147. 
Supplement  of  an  angle,  §  30. 
of  an  arc,  §  190. 
Supplementary-adjacent       angles, 
§33. 
angles,  §  30. 
Surface,  §  2. 


402 


INDEX   TO  DEFINITIONS. 


Surface  of  a  material  body,  §  1. 

of  a  solid,  §  2. 
Symmetrical     polyedral     angles, 
§  455. 
spherical  polygons, 
§591. 

Tangent  circles,  §  150. 
Tetraedron,  §  462. 
Theorem,  §  15. 
Third  proportional,  §  230. 
Transversal,  §  71. 
Trapezium,  §  104. 
Trapezoid,  §  104. 
Triangle,  §  57. 
Triangular  prism,  §  469. 

pyramid,  §  503. 
Triedral  angle,  §  452. 
Tri-rectangular  triangle,  §  698. 
pyramid,  §  629. 
Truncated  prism,  §  472. 

pyramid,  §  505. 

Unit  of  measure,  §  180. 

of  surface,  §  302. 

of  volume,  §  464. 
Upper  base  of  a  frustum  of  a  cone, 
§553. 


Upper  nappe  of  a  conical  sutface, 
§553. 

Variable,  §  184. 
Vertex  of  a  cone,  §  553. 

of  a  conical  surface,  §  553. 
of  a  polyedral  angle,  §  452. 
of  a  pyramid,  §  502. 
.    of    a    spherical    pyramid, 
§  589. 
of  a  triangle,  §  60. 
of  an  angle,  §  20. 
Vertical  angle  of  a  triangle,  §  60. 
angles,  §  28. 
diedral  angles,  §  428. 
polyedral  angles,  §  452. 
Vertices  of  a  polyedron,  §  461. 
of  a  polygon,  §  118. 
of  a  quadrilateral,  §  103. 
of   a  spherical   polygon^ 

§587. 
of  a  triangle,  §  57. 
Volume  of  a  solid,  §  464. 

Zone,  §  662. 

of  one  base,  §  662. 


ANSWERS 


TO 


NUMERICAL   EXERCISES. 


>j»ic 


Book  I. 

4.   24°. 

5.   63°  30',  26°  30'.             8.  22°  30',  157°  30'. 

9.   37°. 

24.   ^  =  112°30',  jB=C=33°45'.         88.   7. 

Book  II. 

12.  28°.      13.   44°  30'.      14.   12°.      15.   54°  30'.      16.   178° 

17.  112°  30'.  18.   83°,  89°  30',  97°,  90°  30',  74°  30'. 

52.  Z  AED  =  14°  30',  Z  AFB  =  10°  30'. 

55.  114°  30',  89°  30',  65°  30',  90°  30'. 

67.  97°30',  89°30',  82°30',  90°30'. 

Book  III. 

1.   112.     2.  42.     3.  If    4.  63.     5.  J5C,  3^,  2f ;  C^,  4,  3; 

AB,  4^,  Sj%.        6.   BC,  llf,  18 j;  CA,  20,  28;  AB,  35,  40. 
7.   19f,  25^.  9.   4  ft.  6  in.  10.   12.  11.   15. 

12.   37  ft.  1  in.  13.   47  ft.  6  in.  14.   -i/VS. 

15.  15V2in.  16.  41.  17.  5^.  18.  21.  19.  24. 
25.  18.  28.  48.  29.  10.  30.  131  31.  9V2.  32.  45. 
34.  17|.  37.  50.  41.  Vl29,  2V21,  V20i.  42.  i^. 
47.    36.         49.   63.         50.   4  and  3 ;  -i/  and  |.         56.   24. 

403 


404  NUMERICAL  EXERCISES. 

57.  17.  58.  21,28.  59.  8V3.  60.  BE  =  4.,  ED  =  12. 
62.   6V3.       67.   14.       70.   21.       74.   ffand^;  9  and  5. 

Book  IV. 
1.  30f  ft.        2.  8  ft.  9  in.        3.  14,  12.       4.  6  ft.  11  in., 
20  ft.  9  in.     5.  6  sq.  ft.  60  sq.  in.     6.   30 V3.     7.  26  yd.  1  ft. 
8.  2  sq.  ft.  48  sq.  in.         9.   243.         10.  210;  24if,  15,  16|. 

11.  73.  12.  117.  16.  2  ft.  10  in.  18.  -2/V3.  19.  3V3. 
21.  120.  24.  210.  25.  18.  26.  li  ft.  27.  6.  28.  4V3. 
29.  1260.  33.  150.  34.  17.  36.  624.  37.  540  sq.  in. 
38.  28  ft.  41.  -1/.  42.  30,16.  45.  it^./47.  AD=^^-\/2, 
AE=llV2.  48.  54.  51.  Area  ^5Z>=39,  area  ^OZ>=45. 
52.  1010.      53.  336. 

Book  V. 

32."  Area, -^I^TT.     33.  Circumference,  34  tt.  34.  64:121. 

36.  9.         37.  13.        38.  |V2.         39.  ^tt.  40.  -^-^-rr. 

41.  9.8268.      42.  J^tt.      43.  392.      44.  48  tt.  45.  1.2732. 

46.    -«/7r.      47.  67r.     48.  167r.     49.  37r,12  7r.     SO.Stt,  87rV2. 

51.9.06.  52.  416  TT  sq.ft.  53.120.99  ft.  54.  57  in. 
60.  67.295''+.      61.  2.658+.      62.  5.64+. 

APPENDIX  TO  PLANE   GEOMETRY. 

58.  10V7.         62.  8.  63.  ff.  91.  480. 

Book  VII. 
1.   4:3.    2.   2:5.    4.   42.    5.    1  ft.  9  in.    6.   ^^\  cu.  in. ; 
63f  sq.  in.       7.   574.       8.   1008.       9.   12,  7.        10.   1944. 

12.  Volume,  50V3.  14.  Volume,  ^f^VS.  15.  17. 
17.  2400  sq.  in.  18.  770.  19.  Volume,  48  V5.  20.  144. 
21.  512,  384.  22.  1705.  23.  10,  1.  24.  36  sq.  in. 
25.   12  in.         28.   V273,  18V237, 180V3.         29.iVll8, 


ANSWERS.  405 

3V109,  15.  30.  V97,  12V93,  72V3.  31.  4V39,  504V3, 
936V3.  32.  6V3,  56V26,  503f  33.  4Vl0,  72V39, 
672V3.  34.  150.  35.  320.  36.  '84(0  37.  700,  1568. 
38.  |V57,  640V3.  39.  42V9i,  624VT40.  108,  21V39. 
41.  240,  J^Vil9.  49.  4V3,  |V2.  50.  15.  51.  768, 
2340.  59.  438.  63.  9600  1b.  64.  50.  69.  168  V3, 
15V219.  76.  3456  cu.  in.  77.  6  ft.  78.  4  ft.  6  in. 
79.  5^4  in.  80.  960,3072.  81.  128.  82.  12.  «3.  6. 
86.  36V3. 


Book  VIII. 

7.  39|.       9.   86°  24'.        10.  36.        11. 

44.        12.   3:2. 

13.  108°.        14.  220.        16.  334.        17. 

66f        18.  36i. 

19.  60.       20.  153°.      23.  45°.       24.  4f 

26.  4V3,  Vs. 

44.  30,8,20. 

Book  IX. 

1.   288  TT,  450  TT,   1296^.  2.   2420.  3.   14,  12. 

4.  300  TT.  5.  2.7489+.  6.  167803.68.  8.  175  tt, 
224  7r,   392  7r.  9..143  7r,   216  TT,   388  tt.  10.   135. 

11.  2800  n-.  12.  1^6  TT.  13.  24.  14.  160  tt,  536  tt. 
15.  4,  1159  77.  16.  24,  260  tt.  17.  128.'  18.  256. 
19.   —in.        20.   7238.2464.         22.   576  tt.         23.   416  tt. 

37r 

24.  130  TT.          25.  2304  TT.          26.  1250^7.          27.   306  tt. 

28.  Volume,   972  tt.                   29.   Area  of  surface,   225  tt. 

32.  347.2956.     '    33.   56°  15'.          34.   58  tt.          37.   81  tt, 

^TT.  38.   8192.              39.   6  in.              40.   1625  oz. 

41.  8  in.            42.   7.            43.   4i.           48.   7rR'(2-V2). 

49.  A^TT.           50.   900  TT,   4500  77.           51.   36  tt  V3,   54  tt. 

53.  32  7rV3.                    55.   720  tt,   3456  tt;    960  tt,   6144  tt. 

56.  5  85  ^^  6 j5  ^.        57.   468  tt.        58.   7rAV2,  ^  ttAVTA. 


406  NUMERICAL  EXERCISES. 

59.   3^  in.,  3^18  in.  60.  ^^~^^^.  61.   ^, 

SV+^D^         go    V9F^  +  3  7rg^F         g3    SWn'L'-S' 
2D      '  '  H  '  '         Stt'L'       ' 

64.  576  7rV2.  65.  ^^'  66.    ^s/SQ^Vi 

GVtt 

67.   !l(^±^,    _Z^!^.  68.   2400.. 

■Vd'  +  6'      3  Va"  4-  b'' 
69.   By  triangle,  ttF,  i  ttA^  ;  by  inscribed  circle,  f  tt^^,  -^  ttJA 

4.Trr'h^         2  TTT^Ji^ 
{2r  +  1if    {2r  +  1if 
72.   2  7ra2 V3,  1  Tra\       73.  67.3698  +  lb.       75.  2  7raV3,  ^ra^. 
76.  -^^TT,  H-TT.  77.   2100  TT.  80.  ^  ^^'^. 

81.   1440 TT.  82.   487.4716.  83.   2Trr'(l+^2\ 

i^r^V2.        86.   96 ,r.      ST  '?JS 

iV7 


«rt        4  7rr-/i"  Z  Trr/i''  iri      o     ^    1        <?    /o 


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Fuller's  Phonetic  Drill  Charts.       Exercises  in  elementary  sounds.    Per  set  (3  charts) 
10  cents. 

Hall's  How  to  Teach  Reading.       Treats  the  important  question  :  what  children  should 
and  should  not  read.     Paper.     25  cts. 

Hyde's  Lessons  in  English,  Book  I.       For  the   lower  grades.      Contains  exercises 
for  reproduction,  picture  lessons,  letter  writing,  uses  of  parts  of  speech,  etc.     35  cts. 

Hyde's  Lessons  in  English,  Book  II.       For  Grammar  schools.     Has  enough  tech- 
nical grammar  for  correct  use  of  language.     50  cts. 

Hyde's  Lessons  in  English,  Book  II  with  Supplement.     Has,  in  addition  to 

the  above,  118  pages  of  technical  grammar.     60  cts.     Supplement  bound  alone,  30  cts. 

Hyde's  Practical  English  Grammar.       For   advanced   classes  in  grammar  .schools 
and  for  high  schools.     50  cts. 

Hyde's  Derivation  of  Words.       With  exercises  on  prefixes,  suffixes,  and  stems.  10  cts. 

Mathews's  Outline  of  English  Grammar,  with   Selections  for  Practice. 

The  application  of  principles  is  made  through  composition  of  original  sentences.     70  cts. 

Penniman's  Common  Words  Difficult  to  Spell.     Graded  list  of  3500  common 

words.     20  cts. 

Penniman's  Prose  Dictation  Exercises.     For  drill  in  spelling,  punctuation  and  use 

of  capitals.     25  cts. 

Phillips's  History  and  Literature  in  Grammar  Grades.     An  essay  showing 

the  intimate  relation  of  the  two  subjects.     15  cts. 

Sever'S  Progressive  Speller.       Gives  spelling,    pronunciation,  definition,  and  use  of 
words.     Vertical  script  is  given  for  script  lessons.     25  cts. 

Smith's  Studies  in  Nature,  and  Language  Lessons.     A  combination  of  object 

lessons  with  language  work.     50  cts.     Part  I  bound  separately,  25  cts. 

Spalding's  Problem  of  Elementary  Composition.     Practical  suggestions  for 

work  in  grammar  grades.     40  cts. 

See  also  our  lists  0/ books  in  Higher  English,  English  Classics, 
Supplementary  Reading,  and  English  Liter/iture, 

D.C.  HEATH  &  CO.,Publishers, Boston,  NewYork,  Chicago 


Wells's    Mathematical    Series. 

ALGEBRA. 

^Vells's  Essentials  of  Algebra  (1897)      ....     $1.10 

A  new  Algebra  for  Secondary  Schools.     With  or  without  answers. 
Wells's  Academic  Algebra  .....       1.08 

This  popular  Algebra  is  intended  for  High  Schools  and  Academies.  It  is  full  in  its 
treatment  of  Factoring  and  contains  an  abundance  of  carefully  selected  problems. 

Wells's  Higher  Algebra    ......       1.32 

The  first  half  of  this  book  is  identical  with  the  corresponding  pages  of  the  Academic 
Algebra.   The  latter  half  treats  more  advanced  topics. 

Wells's  College  Algebra    ......       1.50 

A  thoroughly  modern  text-book  for  colleges  and  scientific  schools.  The  latter  half 
of  this  book,  beginning  with  the  discussion  of  Quadratic  Equations,  is  also  bound 
separately,  and  is  known  as  Wells's  College  Algebra,  Part  II.     $1.32. 

^A^ells's  University  Algebra  .....       1.32 

GEOMETRY. 

Wells's    Essentials  of  Geometry  —  Plane  (1898),  75  cts.;    Solid 

(1899),  75  cts.;  Plane  and  Solid  (1899).  .  .  .1.25 

This  new  text  is  the  latest  in  the  series.  It  offers  a  practical  combination  of  more 
desirable  qualities  than  any  other  Geometry  ever  published. 

Wells's  Elements  of  Geometry  —  Revised  1894. —  Plane,  75  cts.; 

Solid,  75  cts. ;  Plane  and  Solid      .  .  .  .'  .        1.25 

In  his  critical  analysis  of  this  work  the  author  calls  attention  of  instructors  to  some 
forty-nine  points  which  are  worthy  of  special  consideration. 

Wells's  Syllabus  of  Plane  and  Solid  Geometry  .  .         .20 

To  accompany  the  author's  Elements  of  Geometry. 

TRIGONOMETRY. 

Wells's  New  Plane  and  Spherical  Trigonometry  (1896)  .    $1.00 

For  colleges  and  technical  schools.     With  Wells's  New  Six  Place  Tables,  $1.25. 

Wells's  Plane  Trigonometry.       .....         .75 

An  elementary  work  for  secondary  schools.     Contains  Four  Place  Tables. 

Wells's  Essentials  of  Plane  and  Spherical  Trigonometry  .  .90 
For  secondary  schools.  The  chapters  on  Plane  Trigonometry  are  identical  with 
those  of  the  book  described  above  ;  to  these  are  added  three  chapters  on  Spherical 
Trigonometry.     Edition  containing  Tables,  iJi.oS. 

Wells's  New  Six  Place  Logarithmic  Tables     .  .  .60 

The  handsomest  tables  in  print.     Large  page. 

Wells's  Four  Place  Tables .         .  .  .  .  •         .25 

Correspondence  regarding  terms  for  introduction 
and  exchange  is  cordially  invited, 

D.  C.  Heath  &  Co.,  Publishers,  Boston,  New  York,  Chicago 


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SEP     7  19: 


MAR  31 

JAN  30  1935 

APR      1   1S33 

m  26  ^'^ 

REC'D  LD 

AUG  1  2  1963 


19545EP 


3  SENT  ON  ILL 
2  h  2000 


LD  21-50m-l,'3J 


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YB  1 7306 


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